Canadian Mathematics Competitions. Gauss (Grades 7 & 8)

Transcription

1 Canadian Mathematics Competitions Gauss (Grades 7 & 8) s to All Past Problems: Compiled by info@erudits.com.ng

2 The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 015 Gauss Contests (Grades 7 and 8) Wednesday, May 13, 015 (in North America and South America) Thursday, May 14, 015 (outside of North America and South America) s 014 University of Waterloo

3 015 Gauss Contest s Page 3 Grade 7 1. The circle is divided into 4 equal regions. Since 1 of these 4 regions is shaded, then the fraction of the circle that is shaded is 1 4. Answer: (C). Evaluating, 10 (5 ) = 10 3 = 30. Answer: (D) 3. Reading from the graph, Phil ran 4 km, Tom ran 6 km, Pete ran km, Amal ran 8 km, and Sanjay ran 7 km. Therefore, Pete ran the least distance. Answer: (C) 4. The equal-arm balance shows that rectangles have the same mass as 6 circles. If we organize these shapes into two equal piles on both sides of the balance, then we see that 1 rectangle has the same mass as 3 circles. Answer: (B) 5. Of the possible answers, the length of your thumb is closest to 5 cm. Answer: (E) 6. There are 100 centimetres in 1 metre. Therefore, there are = 350 cm in 3.5 metres. Answer: (A) 7. The length of the side not labelled is equal to the sum of the two horizontal lengths that are labelled, or + 3 = 5. Thus, the perimeter of the figure shown is = 0. Answer: (D) 8. The average (mean) number of points scored per game multiplied by the number of games played is equal to the total number of points scored during the season. Therefore, the number of games that Hannah played is equal to the total number of points she scored during the season divided by her average (mean) number of points scored per game, or = 4. Answer: (A) 9. The positive divisors of 0 are: 1,, 4, 5, 10, and 0. Therefore, the number 0 has exactly 6 positive divisors. Answer: (B) 10. Using the digits 4, 7 and 9 without repeating any digit in a given number, the following 3-digit whole numbers can be formed: 479, 497, 749, 794, 947, and 974. There are exactly 6 different 3-digit whole numbers that can be formed in the manner described. Answer: (A) At Gaussville School, 40% or 40 = 4 of the 480 total students voted for math Therefore, the number of students who voted for math is = 4 48 = At Gaussville School, 40% or 0.4 of the 480 total students voted for math. Therefore, the number of students who voted for math is = 19. Answer: (B)

4 015 Gauss Contest s Page 4 1. The first fold creates layers of paper. The second fold places sets of layers together, for a total of 4 layers of paper. Similarly, the third fold places sets of 4 layers of paper together, for a total of 8 layers of paper. That is, each new fold places sets of the previous number of layers together, thereby doubling the previous number of layers. The results of the first five folds are summarized in the table below. Number of folds Number of layers After the sheet has been folded in half five times, the number of layers in the folded sheet is 3. Answer: (B) The multiples of 5 between 1 and 99 are: 5, 10, 15, 0, 5, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95. Of these, only 10, 0, 30, 40, 50, 60, 70, 80, and 90 are even. Therefore, there are 9 even whole numbers between 1 and 99 that are multiples of 5. To create an even multiple of 5, we must multiply 5 by an even whole number (since 5 is odd, multiplying 5 by an odd whole number creates an odd result). The smallest positive even multiple of 5 is 5 = 10. The largest even multiple of 5 less than 99 is 5 18 = 90. That is, multiplying 5 by each of the even numbers from to 18 results in the only even multiples of 5 between 1 and 99. Since there are 9 even numbers from to 18 (inclusive), then there are 9 even whole numbers between 1 and 99 that are multiples of 5. Answer: (C) 14. Consider the value of U in the diagram shown. Since a 3 already occurs in the second row, then U cannot equal 3 (each of the numbers 1,, 3, occur only once in each row). Since a 1 already occurs in the third column, then U cannot equal 1 (each of the numbers 1,, 3, occur only once in each column). Since U cannot equal 3 or 1, then U =. Therefore, a and a 3 already occur in the second row and so X = 1. At this point, a and a 1 already occur in the third column and so Y = 3. The value of X + Y = = 4. 3 X 1 U Y Answer: (E) 15. The rectangle has area 5 1 = 60 cm. Each of the two congruent unshaded triangles has area 1 5 = 5 cm. The area of the shaded region is equal to the area of the rectangle minus the areas of the two unshaded triangles, which is = 50 cm. Answer: (E)

5 015 Gauss Contest s Page The total value of one quarter, one dime and one nickel is = 40. Since you have equal numbers of quarters, dimes and nickels, you can separate your coins into piles, each containing exactly 1 quarter, 1 dime, and 1 nickel. Each pile has a value of 40, and since = 11, then you must have 11 quarters, 11 dimes and 11 nickels. Therefore, you have 11 dimes. Note: You can check that 11 ( ) = = 440 = $4.40, as required. Answer: (B) 17. The original cube (before the corner was cut off) had 1 edges. Cutting off the corner does not eliminate any of the 1 edges of the original cube. However, cutting off the corner does add 3 edges that were not present originally, the 3 edges of the new triangular face. Since no edges of the original cube were lost, but 3 new edges were created, then the new solid has = 15 edges. Answer: (D) 18. To find the image of P Q, we reflect points P and Q across the x-axis, then join them. Since P is 3 units above the x-axis, then the reflection of P across the x-axis is 3 units below the x-axis at the same x-coordinate. That is, point T is the image of P after it is reflected across the x-axis. Similarly, after a reflection across the x-axis, the image of point Q will be 6 units below the x-axis but have the same x-coordinate as Q. That is, point U is the image of Q after it is reflected across the x-axis. Therefore, the line segment T U is the image of P Q after it is reflected across the x-axis. Answer: (B) 19. Since the number of digits that repeat is 6, then the digits begin to repeat again after 10 digits (since 10 = 6 0). That is, the 11 st digit is a 1, the 1 nd digit is a 4, and the 13 rd digit is a. Answer: (C) 0. Since the sum of the measures of the three angles in any triangle is 180, then the sum of the measures of the two unknown angles in the given triangle is = 135. The measures of the two unknown angles are in the ratio 4 : 5, and so one of the two angle 5 measures is = 5 of the sum of the two angles, while the other angle measures 4 = 4 of the sum of the two angles. That is, the larger of the two unknown angles measures = 75, and the smaller of the unknown angles measures = 60. We may check that = 180. The largest angle in the triangle measures 75. Answer: (C) 1. We begin by choosing the largest number in each row, 5, 10, 15, 0, 5, and calling this list L. The sum of the five numbers in L is = 75 and this sum satisfies the condition that no two numbers come from the same row. However, the numbers in L are taken from columns 1 and 5 only, and the numbers must be chosen so that no two come from the same column. Thus, the largest of the five answers given, 75, is not possible. Note: In assuring that we take one number from each row, this choice of numbers, L, is the only way to obtain a sum of 75 (since we chose the largest number in each row).

6 015 Gauss Contest s Page 6 Of the five answers given, the next largest answer is 73. Since L uses the largest number in each row and has a sum of 75, we can obtain a sum of 73 either by replacing one of the numbers in L with a number that is two less, or by replacing two of the numbers in L with numbers that are each one less. For example, the list 3, 10, 15, 0, 5 (one change to L) has sum 73 as does the list 4, 9, 15, 0, 5 (two changes to L). That is, to obtain a sum of 73 while choosing exactly one number from each row, we must choose at least three of the numbers from L. However, since two numbers in L lie in column 1 and three numbers from L lie in column 5, it is not possible to choose at least three numbers from L so that no two of the numbers are from the same column. Any other replacement would give a sum less than 73, which would require the replacement of a number with a larger number in another row to compensate. This is impossible since each row is represented in L by the largest number in the row. Therefore, it is not possible to obtain a sum of 73. Of the five answers given, the next largest answer is 71. By choosing the numbers, 3, 9, 14, 0, 5 we obtain the sum = 71 while satisfying the condition that no two numbers come from the same row and no two numbers come from the same column. Thus, 71 is the largest possible sum that satisfies the given conditions. Note: There are other choices of five numbers which also give a sum of 71 and satisfy the given conditions. Answer: (C). Since the perimeter of the square is P and the 4 sides of a square are equal in length, then each side of the square has length 1 P. We now work backward to determine the width and length 4 of the rectangle. The width of the rectangle was doubled to produced the side of the square with length 1P. 4 Therefore, the width of the rectangle is half of the side length of the square, or P = 1 8 P. The length of the rectangle was halved to produce the side of the square with length 1 4 P. Therefore, the length of the rectangle is twice the side length of the square, or 1P = 1P. 4 Finally, we determine the perimeter of the rectangle having width 1P and length 1P, obtaining 8 ( 1 P + 1P ) = ( 1 P + 4P ) = ( 5 P ) = 5P Answer: (D) 3. 1 Every 4-digit palindrome is of the form abba, where a is a digit between 1 and 9 inclusive and b is a digit between 0 and 9 inclusive (and b is not necessarily different than a). Every 5-digit palindrome is of the form abcba, where a is a digit between 1 and 9 inclusive, b is a digit between 0 and 9 inclusive (and b is not necessarily different than a), and c is a digit between 0 and 9 inclusive (and c is not necessarily different than a and b). That is, for every 4-digit palindrome abba there are 10 possible digits c so that abcba is a 5-digit palindrome. For example if a = and c = 3, then the 4-digit palindrome 33 can be used to create the 10 5-digit palindromes: 303, 313, 33, 333, 343, 353, 363, 373, 383, 393. Thus, for every 4-digit palindrome abba, there are exactly 10 5-digit palindromes abcba and so the ratio of the number of 4-digit palindromes to the number of 5-digit palindromes is 1 : 10.

7 015 Gauss Contest s Page 7 Every 4-digit palindrome is of the form abba, where a is a digit between 1 and 9 inclusive and b is a digit between 0 and 9 inclusive (and b is not necessarily different than a). There are 9 choices for the first digit a and, for each of these choices, there are 10 choices for the second digit b or 9 10 = 90 choices for the first two digits ab. Once the first two digits of the 4-digit palindrome are chosen, then the third and fourth digits are also determined (since the third digit must equal the second and the fourth must equal the first). That is, there are 90 4-digit palindromes. Every 5-digit palindrome is of the form defed, where d is a digit from 1 to 9 inclusive and e is a digit from 0 to 9 inclusive (and e is not necessarily different than d) and f is a digit from 0 to 9 inclusive (and f is not necessarily different than d and e). There are 9 choices for the first digit d and 10 choices for the second digit e and 10 choices for the third digit f or = 900 choices for the first three digits def. Once the first three digits of the 5-digit palindrome are chosen, then the fourth and fifth digits are also determined (since the fourth digit must equal the second and the fifth must equal the first). That is, there are digit palindromes. Thus, the ratio of the number of 4-digit palindromes to the number of 5-digit palindromes is 90 : 900 or 1 : 10. Answer: (E) 4. We can determine which triangle has the greatest area by using a fixed side length of 4 for each of the identical squares and using this to calculate the unknown areas. Q We begin by constructing P V U and noticing that it is contained V W A R within square QABP, as shown. The area of P V U is X determined by subtracting the areas of triangles P QV, V AU and Y P BU from the area of square QABP. U Z Since QA = 8 and AB = 8, then the area of square QABP is P B S 8 8 = 64. Since P Q = 8 and QV =, then the area of P QV is 1 8 = 8. Since V A = 6 and AU = 6, then the area of V AU is = 18. Since P B = 8 and UB =, then the area of P BU is 1 8 = 8. Therefore, the area of P V U is = 30. Next, we construct P XZ and then construct rectangle CDSP by drawing CD parallel to P S through X. Further, X is the midpoint of the side of a square and so C and D are also midpoints of the sides of their respective squares. P S The area of P XZ is determined by subtracting the areas of triangles P CX, XDZ and P SZ from the area of rectangle CDSP. Since CD = 1 and DS = 6, then the area of rectangle CDSP is 1 6 = 7. Since P C = 6 and CX = 8, then the area of P CX is = 4. Since XD = 4 and DZ = 4, then the area of XDZ is = 8. Since P S = 1 and ZS =, then the area of P SZ is 1 1 = 1. Therefore, the area of P XZ is = 8. Q C V W U X Y R D Z

8 015 Gauss Contest s Page 8 Q Construct 4P V X and notice that it is contained within square QABP, as shown. The area of 4P V X is determined by subtracting the areas of triangles P QV, V AX and P BX from the area of square QABP. As we previously determined, the area of square QABP is 64 and P the area of 4P QV is 8. 1 Since V A = 6 and AX =, then the area of 4V AX is 6 = 6. Since P B = 8 and XB = 6, then the area of 4P BX is = 4. Therefore, the area of 4P V X is = 6. V Q V Construct 4P Y S and the perpendicular from Y to E on P S, as shown. Since P S = 1 and Y E = 4 (Y E is parallel to RS and thus equal in length to the side of the square), then the area of 4P Y S is = 4. Construct 4P QW, as shown. Since P Q = 8 and QW = 6, then the area of 4P QW is = 4. The areas of the 5 triangles are 30, 8, 6, 4, and 4. The triangle with greatest area, 30, is 4P V U. W A R X Y U S B W R X Y Z U P Q V Z E W S R X U P Y Z S Answer: (A) 5. All -digit prime numbers are odd numbers, so to create a reversal pair, both digits of each prime must be odd (so that both the original number and its reversal are odd numbers). We also note that the digit 5 cannot appear in either prime number of the reversal pair since any -digit number ending in 5 is not prime. Combining these two facts together leaves only the following list of prime numbers from which to search for reversal pairs: 11, 13, 17, 19, 31, 37, 71, 73, 79, and 97. This allows us to determine that the only reversal pairs are: 13 and 31, 17 and 71, 37 and 73, and 79 and 97. (Note that the reversal of 11 does not produce a different prime number and the reversal of 19 is 91, which is not prime since 7 13 = 91.) Given a reversal pair, we must determine the prime numbers (different than each prime of the reversal pair) whose product with the reversal pair is a positive integer less than The product of the reversal pair 79 and 97 is = Since the smallest prime number is and 7663 = 15 36, which is greater than , then the reversal pair 79 and 97 gives no possibilities that satisfy the given conditions. We continue in this way, analyzing the other 3 reversal pairs, and summarize our results in the table below.

9 015 Gauss Contest s Page 9 Prime Product of the Prime Number with the Reversal Pair Number 13 and and and and = = = 540 greater than = = = = = 6035 greater than = = = 4433 greater than can t use 13 twice = = = greater than Total In any column, once we obtain a product that is greater than , we may stop evaluating subsequent products since they use a larger prime number and thus will exceed the previous product. In total, there are = 14 positive integers less than which have the required property. Answer: (B)

10 015 Gauss Contest s Page 10 Grade 8 1. Evaluating, = = = Since there are 60 minutes in an hour, then 40 minutes after 10:0 it is 11:00. Therefore, 45 minutes after 10:0 it is 11: Of the possible answers, the length of your thumb is closest to 5 cm. Answer: (A) Answer: (E) Answer: (E) 4. Reading from the graph, Phil ran 4 km, Tom ran 6 km, Pete ran km, Amal ran 8 km, and Sanjay ran 7 km. Ordering these distances from least to greatest, we get Pete ran km, Phil ran 4 km, Tom ran 6 km, Sanjay ran 7 km, and Amal ran 8 km. In this ordered list of 5 distances, the median distance is in the middle, the third greatest. Therefore, Tom ran the median distance. Answer: (B) 5. 1 Since x + 3 = 10, then x = 10 3 = 7. When x = 7, the value of 5x + 15 is 5(7) + 15 = = 50. When multiplying x + 3 by 5, we get 5 (x + 3) = 5 x = 5x Since x + 3 = 10, then 5 (x + 3) = 5 10 = 50. Therefore, 5 (x + 3) = 5x + 15 = 50. Answer: (E) 6. The two equal widths, each of length 4, contribute 4 = 8 to the perimeter of the rectangle. The two lengths contribute the remaining 4 8 = 34 to the perimeter. Since the two lengths are equal, they each contribute 34 = 17 to the perimeter. Therefore, the length of the rectangle is 17. Answer: (B) 7. To begin, there are 4 circles and rectangles on the left arm, balanced by 10 circles on the right arm. If we remove 4 circles from each side of the equal-arm scale, the scale will remain balanced (since we are removing the same mass from each side). That is, the rectangles that will remain on the left arm are equal in mass to the 6 circles that will remain on the right arm. Since rectangles are equal in mass to 6 circles, then 1 rectangle has the same mass as 3 circles. Answer: (B) 8. 1 A 5% increase in 160 is equal to or = Therefore, Aidan s height increased by 8 cm over the summer. His height at the end of the summer was = 168 cm. Since Aidan s 160 cm height increased by 5%, then his height at the end of the summer was ( ) 160 = ( ) 160 = = 168 cm. Answer: (A)

11 015 Gauss Contest s Page When x = 4 and y =, x + y = 4 + = 6, xy = 4 = 8, x y = 4 =, x y = 4 =, and y x = 4 = 1. Therefore, the expression which gives the smallest value when x = 4 and y = is y x. Answer: (E) Evaluating using a denominator of 1, = 6 4 is 9. Since = + 1 = 3 and 3 = 9, then 1 + = The number represented by is = Straight angles measure 180. Therefore, y = 180, and so y = = 40. The three interior angles of any triangle add to 180. Thus, z = 180, and so z = = 60. Opposite angles have equal measures. Since the angle measuring z is opposite the angle measuring x, then x = z = and so the number represented by 80º yº 140º zº Answer: (C) xº Answer: (C) 1. Since Zara s bicycle tire has a circumference of 1.5 m, then each full rotation of the tire moves the bike 1.5 m forward. If Zara travels 900 m on her bike, then her tire will make = 600 full rotations. Answer: (C) 13. To find the image of P Q, we reflect points P and Q across the x-axis, then join them. Since P is 3 units above the x-axis, then the reflection of P across the x-axis is 3 units below the x-axis at the same x-coordinate. That is, point T is the image of P after it is reflected across the x-axis. Similarly, after a reflection across the x-axis, the image of point Q will be 6 units below the x-axis but have the same x-coordinate as Q. That is, point U is the image of Q after it is reflected across the x-axis. Therefore, the line segment T U is the image of P Q after it is reflected across the x-axis. Answer: (B) 14. In the table below, we determine the total value of the three bills that remain in Carolyn s wallet when each of the four bills is removed. Bill Removed Sum of the Bills Remaining $5 $10+$0+$50=$80 $10 $5+$0+$50=$75 $0 $5+$10+$50=$65 $50 $5+$10+$0=$35 It is equally likely that any one of the four bills is removed from the wallet and therefore any of the four sums of the bills remaining in the wallet is equally likely. Of the four possible sums, $80, $75, $65, and $35, two are greater than $70. Therefore, the probability that the total value of the three bills left in Carolyn s wallet is greater than $70, is or Answer: (A)

12 015 Gauss Contest s Page In the table below, we list the mass of each dog at the end of each month. Month Walter s mass (in kg) Stanley s mass (in kg) After 1 months have passed, Stanley s mass is 36 kg and is equal to Walter s mass. (Note that since Stanley s mass is increasing at a greater rate than Walter s each month, this is the only time that the two dogs will have the same mass.) Answer: (D) 16. First, we must determine the perimeter of the given triangle. Let the unknown side length measure x cm. Since the triangle is a right-angled triangle, then by the Pythagorean Theorem we get x = or x = = 100 and so x = 100 = 10 (since x > 0). Therefore the perimeter of the triangle is = 4 cm and so the perimeter of the square is also 4 cm. Since the 4 sides of the square are equal in length, then each measures 4 = 6 cm. 4 Thus, the area of the square is 6 6 = 36 cm. Answer: (D) 17. Since the number of digits that repeat is 6, then the digits begin to repeat again after 10 digits (since 10 = 6 0). That is, the 11 st digit is a 1, the 1 nd digit is a 4, and the 13 rd digit is a. Answer: (C) 18. Using the definition of, we see that p 3 = p 3 + p + 3 = 3p + p + 3 = 4p + 3. Since p 3 = 39, then 4p + 3 = 39 or 4p = 39 3 = 36 and so p = 36 4 = 9. Answer: (C) Originally there are 3 times as many boys as girls, so then for every 3 boys there is 1 girl and = 4 children in the room. That is, the number of boys in the room is 3 of the number of children in the room. 4 Next we consider each of the 5 possible answers, in turn, to determine which represents the total number of children in the room originally. If the original number of children in the room is 15 (as in answer (A)), the number of boys is 3 15 = 45 = Since it is not possible to have 11.5 boys in the room, then we know that 15 is not the correct answer. If the original number of children in the room is 0 (as in answer (B)), the number of boys is 3 0 = 60 = If the number of boys in the room was originally 15, then the number of girls was 0 15 = 5. Next we must check if there will be 5 times as many boys as girls in the room once 4 boys and 4 girls leave the room. If 4 boys leave the room, there are 11 boys remaining. If 4 girls leave the room, there is 1 girl remaining and since there are not 5 times as many boys as girls, then 0 is not the correct answer. If the original number of children in the room is 4 (as in answer (C)), the number of boys is 3 4 = 7 = If the number of boys in the room was originally 18, then the number of girls was 4 18 = 6.

13 015 Gauss Contest s Page 13 If 4 boys leave the room, there are 14 boys left and if 4 girls leave the room, then there are girls left. Since there are not 5 times as many boys as girls, then 4 is not the correct answer. If the original number of children in the room is 3 (as in answer (D)), the number of boys is = 96 4 = 4. If the number of boys in the room was originally 4, then the number of girls was 3 4 = 8. If 4 boys leave the room, there are 0 left and if 4 girls leave the room, then there are 4 left. Since there are 5 times as many boys as girls, then we know that the original number of children is 3. (Note: We may check that the final answer, 40, gives 30 boys and 10 girls originally and when 4 boys and 4 girls leave the room there are 6 boys and 6 girls which again does not represent 5 times as many boys as girls.) Originally there are 3 times as many boys as girls, so if there are x girls in the room, then there are 3x boys. If 4 boys leave the room, there are 3x 4 boys remaining. If 4 girls leave the room, there are x 4 girls remaining. At this point, there are 5 times as many boys as girls in the room. That is, 5 (x 4) = 3x 4 or 5x 0 = 3x 4 and so 5x 3x = 0 4 or x = 16 and so x = 8. Therefore, the original number of girls in the room is 8 and the original number of boys is 3 8 = 4. The original number of students in the room is = 3. Answer: (D) 0. 1 Call the given vertex of the rectangle (1, ) point X and name each of the 5 answers to match the letters A through E, as shown. Point E(1, 1) is 3 units below point X (since their x-coordinates are equal and their y-coordinates differ by 3). Thus, E(1, 1) could be the coordinates of a vertex of a 3 by 4 rectangle having vertex X (X and E would be adjacent vertices of the rectangle). Point C(5, 1) is 3 units below and 4 units right of point X (since their y-coordinates differ by 3 and their x-coordinates differ by 4). Thus, C(5, 1) could be the coordinates of a vertex of a 3 by 4 rectangle having vertex X (X and C would be opposite vertices of the rectangle). D(-,6) A(-3,-1) y X(1,) x C(5,-1) E(1,-1) B(1,-5) Point A( 3, 1) is 3 units below and 4 units left of point X (since their y-coordinates differ by 3 and their x-coordinates differ by 4). Thus, A( 3, 1) could be the coordinates of a vertex of a 3 by 4 rectangle having vertex X (X and A would be opposite vertices of the rectangle). Point D(, 6) is 4 units above and 3 units left of point X (since their y-coordinates differ by 4 and their x-coordinates differ by 3). Thus, D(, 6) could be the coordinates of a vertex of a 3 by 4 rectangle having vertex X (X and D would be opposite vertices of the rectangle). The only point remaining is B(1, 5) and since it is possible for each of the other 4 answers to

14 015 Gauss Contest s Page 14 be one of the other vertices of the rectangle, then it must be (1, 5) that can not be. Point B(1, 5) is 7 units below point X (since their y-coordinates differ by 7). How might we show that no two vertices of a 3 by 4 rectangle are 7 units apart? (See ). The distance between any two adjacent vertices of a 3 by 4 rectangle P QRS is either 3 or 4 units (such as P and Q or Q and R, as shown). P S The distance between any two opposite vertices of a rectangle (such as P and R) can be found using the Pythagorean Theorem. 3 In the right-angled triangle P QR, we get P R = or P R = = 5 and so P R = Q 4 R 5 = 5 (since P R > 0). That is, the greatest distance between any two vertices of a 3 by 4 rectangle is 5 units. As shown and explained in 1, the distance between X(1, ) and B(1, 5) is 7 units. Therefore (1, 5) could not be the coordinates of one of the other vertices of a 3 by 4 rectangle having vertex X(1, ). Answer: (B) 1. In square P QRS, P S = SR and since M and N are midpoints of these sides having equal length, then MS = SN. The area of SMN is 1 MS SN. Since this area equals 18, then 1 MS SN = 18 or MS SN = 36 and so MS = SN = 6 (since they are equal in length). The side of the square, P S, is equal in length to P M + MS = = 1 (since M is the midpoint of P S) and so P S = SR = RQ = QP = 1. The area of QMN is equal to the area of square P QRS minus the combined areas of the three right-angled triangles, SMN, NRQ and QP M. Square P QRS has area P S SR = 1 1 = 144. SMN has area 18, as was given in the question. NRQ has area 1 QR RN = = 36 (since SN = RN = 6). QP M has area 1 QP P M = = 36. Thus the area of QMN is = 54. Answer: (E). Let the number of adult tickets sold be a. Since the price for each adult ticket is $1, then the revenue from all adult tickets sold (in dollars) is 1 a or 1a. Since the number of child tickets sold is equal to the number of adult tickets sold, we can let the number of child tickets sold be a, and the total revenue from all $6 child tickets be 6a (in dollars). In dollars, the combined revenue of all adult tickets and all child tickets is 1a + 6a = 18a. Since the total number of tickets sold is 10, and a adult tickets were sold and a child tickets were sold, then the remaining 10 a tickets were sold to seniors. Since the price for each senior ticket is $10, then the revenue from all senior tickets sold (in dollars) is 10 (10 a). Thus the combined revenue from all ticket sales is 10 (10 a) + 18a, dollars.

15 015 Gauss Contest s Page 15 The total revenue from the ticket sales was $1100 and so 10 (10 a) + 18a = Solving this equation, we get a + 18a = 1100 or 100 0a + 18a = 1100 or 100 a = 1100 and so a = 100 or a = 50. Therefore, the number of senior tickets sold for the concert was 10 a = 10 (50) = = 0. We may check that the number of tickets sold to each of the three groups gives the correct total revenue. Since the number of adult tickets sold was equal to the number of child tickets sold which was equal to a, then 50 of each were sold. The revenue from 50 adult tickets is 50 $1 = $600. The revenue from 50 child tickets is 50 $6 = $300. The revenue from 0 senior tickets is 0 $10 = $00. The total revenue from all tickets sold was $600 + $300 + $00 = $1100, as required. Answer: (B) 3. The list of integers 4, 4, x, y, 13 has been arranged from least to greatest, and so 4 x and x y and y 13. The sum of the 5 integers is x + y + 13 = 1 + x + y and so the average is 1 + x + y. 5 Since this average is a whole number, then 1 + x + y must be divisible by 5 (that is, 1 + x + y is a multiple of 5). How small and how large can the sum 1 + x + y be? We know that 4 x and x y, so the smallest that x + y can be is = 8. Since x + y is at least 8, then 1 + x + y is at least = 9. Using the fact that x y and y 13, the largest that x + y can be is = 6. Since x + y is at most 6, then 1 + x + y is at most = 47. The multiples of 5 between 9 and 47 are 30, 35, 40, and 45. When 1 + x + y = 30, we get x + y = 30 1 = 9. The only ordered pair (x, y) such that 4 x and x y and y 13, and x + y = 9 is (x, y) = (4, 5). Continuing in this way, we determine all possible values of x and y that satisfy the given conditions in the table below. Value of 1 + x + y Value of x + y Ordered Pairs (x, y) with 4 x and x y and y = 9 (4, 5) = 14 (4, 10), (5, 9), (6, 8), (7, 7) = 19 (6, 13), (7, 1), (8, 11), (9, 10) = 4 (11, 13), (1, 1) The number of ordered pairs (x, y) such that the average of the 5 integers 4, 4, x, y, 13 is itself an integer is 11. Answer: (E) 4. The two joggers meet every 36 seconds. Therefore, the combined distance that the two joggers run every 36 seconds is equal to the total distance around one lap of the oval track, which is constant. Thus the greater the first jogger s constant speed, the greater the distance that they run every 36 seconds, meaning the second jogger runs less distance in the same time (their combined

16 015 Gauss Contest s Page 16 distance is constant) and hence the smaller the second jogger s constant speed. Conversely, the slower the first jogger s constant speed, the less distance that they run every 36 seconds, meaning the second jogger must run a greater distance in this same time and hence the greater the second jogger s constant speed. This tells us that if the first jogger completes one lap of the track as fast as possible, which is in 80 seconds, then the second jogger s time to complete one lap of the track is as slow as possible. We will call this time t max, the maximum possible time that it takes the second jogger to complete one lap of the track. Similarly, if the first jogger completes one lap of the track as slowly as possible, which is in 100 seconds, then the second jogger s time to complete one lap of the track is as fast as possible. We will call this time t min, the minimum possible time that it takes the second jogger to complete one lap of the track. Finding the value of t max Recall that t max is the time it takes the second jogger to complete one lap when the first jogger completes one lap of the track in 80 seconds. If the first jogger can complete one lap of the track in 80 seconds, then in 36 seconds of running, the first jogger will complete = 9 of a complete lap of the track. 0 In this same 36 seconds, the two joggers combined distance running is 1 lap, and so the second jogger runs = 11 of a complete lap. 0 If the second jogger runs 11 of a complete lap in 36 seconds, then the second jogger runs = 1 complete lap in 36 = seconds. Thus, t max = = seconds. 0 Finding the value of t min Recall that t min is the time it takes the second jogger to complete one lap when the first jogger completes one lap of the track in 100 seconds. If the first jogger can complete one lap of the track in 100 seconds, then in 36 seconds of running, the first jogger will complete 36 of a complete lap of the track. 100 = 9 5 In this same 36 seconds, the two joggers combined distance running is 1 lap, and so the second jogger runs = 16 of a complete lap. 5 If the second jogger runs 16 of a complete lap in 36 seconds, then the second jogger runs = 1 complete lap in 36 = seconds. Thus, t min = 900 = 56.5 seconds. 16 Determining the product of the smallest and largest integer values of t Since the second jogger completes 1 lap of the track in at most seconds, then the largest possible integer value of t is 65 seconds. The second jogger completes 1 lap of the track in at least 56.5 seconds, so then the smallest possible integer value of t is 57 seconds. Finally, the product of the smallest and largest integer values of t is = Answer: (A)

17 015 Gauss Contest s Page Let the alternating sum of the digits be S. If the 7-digit integer is abcdefg, then S = a b + c d + e f + g. This sum can be grouped into the digits which contribute positively to the sum, and those which contribute negatively to the sum. Rewriting the sum in this way, we get S = (a + c + e + g) (b + d + f). Taking the 4 digits which contribute positively to S (there are always 4), we let P = a+c+e+g. Similarly, taking the 3 digits which contribute negatively to S (there are always 3), we let N = b + d + f. Thus, it follows that S = (a + c + e + g) (b + d + f) = P N. We determine the largest possible value of S by choosing the 4 largest integers, 4, 5, 6, 7 (in any order), to make up P, and choosing the 3 smallest integers, 1,, 3 (in any order), to make up N. That is, the largest possible alternating sum is S = ( ) ( ) = 16. We determine the smallest possible value of S by choosing the 4 smallest integers, 1,, 3, 4 (in any order), to make up P, and choosing the 3 largest integers, 5, 6, 7 (in any order), to make up N. That is, the smallest possible alternating sum is S = ( ) ( ) = 8. Since S must be divisible by 11 (with S 8 and S 16), then either S = 11 or S = 0. The sum of the first 7 positive integers is = 8, and since each of these 7 integers must contribute to either P or to N, then P + N = 8. Case 1: The alternating sum of the digits is 11, or S = 11 If S = 11, then S = P N = 11. If P N is 11 (an odd number), then either P is an even number and N is odd, or the opposite is true (they can t both be odd and they can t both be even). That is, the difference between two integers is odd only if one of the integers is even and the other is odd (we say that P and N have different parity). However, if one of P or N is even and the other is odd, then their sum P + N is also odd. But we know that P + N = 8, an even number. Therefore, it is not possible that S = 11. There are no 7-digit integers formed from the integers 1 through 7 that have an alternating digit sum of 11 and are divisible by 11. Case : The alternating sum of the digits is 0, or S = 0 If S = 0, then S = P N = 0 and so P = N. Since P + N = 8, then P = N = 14. We find all groups of 3 digits, chosen from the digits 1 to 7, such that their sum N = 14. There are exactly 4 possibilities: (7, 6, 1), (7, 5, ), (7, 4, 3), and (6, 5, 3). In each of these 4 cases, the digits from 1 to 7 that were not chosen, (, 3, 4, 5), (1, 3, 4, 6), (1,, 5, 6), and (1,, 4, 7), respectively, represent the 4 digits whose sum is P = 14. We summarize this in the table below. 4 digits whose 3 digits whose examples of 7-digit sum is P = 14 sum is N = 14 integers created from these, 3, 4, 5 7, 6, , , 3, 4, 6 7, 5, , ,, 5, 6 7, 4, , ,, 4, 7 6, 5, ,

18 015 Gauss Contest s Page 18 Consider the first row of numbers in this table above. Each arrangement of the 4 digits, 3, 4, 5 combined with each arrangement of the 3 digits 7, 6, 1 (in the required way) gives a new 7-digit integer whose alternating digit sum is 0. Two such arrangements are shown (you may check that S = 0 for each). Since there are = 4 ways to arrange the 4 digits (4 choices for the first digit, 3 choices for the second, choices for the third and 1 choice for the last digit), and 3 1 = 6 ways to arrange the 3 digits, then there are 4 6 = 144 ways to arrange the 4 digits and the 3 digits. Each of these 144 arrangements is different from the others, and since P = N = 14 for each, then S = P N = 0 and so each of the digit numbers is divisible by 11. Similarly, there are also 144 arrangements that can be formed with each of the other 3 groups of integers that are shown in the final 3 rows of the table. That is, there are a total of = digit integers (formed from the integers 1 through 7) which are divisible by 11. The total number of 7-digit integers that can be formed from the integers 1 through 7 is equal to the total number of arrangements of the integers 1 through 7, or = Therefore, when the digits 1 through 7 are each used to form a random 7-digit integer, the probability that the number formed is divisible by 11 is = Answer: (E)

19 The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 014 Gauss Contests (Grades 7 and 8) Wednesday, May 14, 014 (in North America and South America) Thursday, May 15, 014 (outside of North America and South America) s 013 University of Waterloo

20 014 Gauss Contest s Page 3 Grade 7 1. Evaluating, (4 3) + = 1 + = 14. Answer: (C). 1 We place each of the five answers and 100 on a number line. Of the five answers given, the two closest numbers to 100 are 98 and 103. Since 98 is units away from 100 and 103 is 3 units away from 100, then 98 is closest to We calculate the positive difference between 100 and each of the five possible answers. The number closest to 100 on the number line will produce the smallest of these positive differences. Possible Answers Positive = = = = = 10 Differences Since the smallest positive difference is, then 98 is the closest to 100 on the number line. Answer: (A) 3. Since five times the number equals one hundred, then the number equals one hundred divided by five. Therefore the number is = 0. Answer: (E) 4. The spinner has 6 sections in total and of these sections contain the letter P. Sections are equal to one another in size and thus they are each equally likely to be landed on. Therefore, the probability of landing on a section that contains the letter P is 6 Ȧnswer: (D) 5. Each scoop of fish food can feed 8 goldfish. Therefore, 4 scoops of fish food can feed 4 8 = 3 goldfish. 6. Both the numerator and the denominator are divisible by 5. Dividing, we get 15 = 15 5 = Answer: (E). Therefore, 15 5 is equivalent to 3 5. Answer: (C) 7. The largest two-digit number that is a multiple of 7 is 7 14 = 98. Thus, there are 14 positive multiples of 7 that are less than 100. However, this includes 7 1 = 7 which is not a two-digit number. Therefore, there are 14 1 = 13 positive two-digit numbers which are divisible by 7. (Note that these 13 numbers are 14, 1, 8, 35, 4, 49, 56, 63, 70, 77, 84, 91, and 98.) Answer: (E) 8. 1 Evaluating the left side of the equation, we get = 86. Therefore the right side of the equation, 10, must also equal 86. Since = 86, then the value represented by the is 14. Since = ( ) ( ) = = 10 14, then the value represented by the is 14. Answer: (D)

21 014 Gauss Contest s Page 4 9. The measure of P ZQ formed by the bottom edge of the shaded quadrilateral, ZQ, and this same edge after the rotation, ZP, is approximately 90. The transformation of the shaded quadrilateral to the unshaded quadrilateral is a clockwise rotation around point Z through an angle equal to the measure of reflex angle P ZQ. The measure of P ZQ added to the measure of reflex P ZQ is equal to the measure of one complete rotation, or 360. Therefore, the measure of reflex angle P ZQ is approximately or 70. Thus the clockwise rotation around point Z is through an angle of approximately 70. P Z Q Answer: (B) 10. In the table below, each of the five expressions is evaluated using the correct order of operations. Expression (A) Value = = 11 (B) = = (C) = 3 6 = 17 (D) = = = 1 4 (E) = = = 8 5 The only expression that is equal to 17 is , or (C). Answer: (C) 11. Since each of the numbers in the set is between 0 and 1, then the tenths digit of each number contributes more to its value than any of its other digits. The largest tenths digit of the given numbers is 4, and so 0.43 is the largest number in the set. The smallest tenths digit of the given numbers is 0, so is the smallest number in the set. Therefore, the sum of the smallest number in the set and the largest number in the set is = Answer: (D) 1. The two diagonals of a square bisect one another (divide each other into two equal lengths) at the centre of the square. Therefore, the two diagonals divide the square into four identical triangles. One of these four triangles is the shaded region which has area equal to one quarter of the area of the square. Since the area of the square is 8 8 = 64 cm, the area of the shaded region is 64 4 = 16 cm. Answer: (C) 13. The sum of the three numbers in the first column is = 36. The sum of the numbers in each column and in each row in the square is the same and so the sum of the three numbers in the second row is also 36. That is, 14 + x + 10 = 36 or x + 4 = 36, and so x = 36 4 = 1. Answer: (E)

22 014 Gauss Contest s Page We systematically count rectangles by searching for groups of rectangles that are of similar size. The largest rectangles in the diagram are all roughly the same size and overlap in pairs. There are 3 of these; each is shaded black and shown below. Rectangle 1 Rectangle Rectangle 3 Rectangle (shown above) consists of 4 small rectangles. We shade these rectangles black and label them 4, 5, 6, 7, as shown below. Rectangle 4 Rectangle 5 Rectangle 6 Rectangle 7 Together, Rectangle 4 and Rectangle 5 (shown above) create Rectangle 8, shown below. Similarly, Rectangle 6 and Rectangle 7 together create Rectangle 9, shown below. Rectangle 8 Rectangle 9 Finally, Rectangle 4 and Rectangle 6 (shown above) together create Rectangle 10, shown below. Similarly, Rectangle 5 and Rectangle 7 together create Rectangle 11, shown below. Rectangle 10 Rectangle 11 There are no other rectangles that can be formed. In total, there are 11 rectangles in the given diagram. Answer: (A) 15. The horizontal translation needed to get from Lori s house to Alex s house is the difference between the x-coordinate of Lori s house, 6, and the x-coordinate of Alex s house,, or 6 ( ) = 6 + = 8. The vertical translation needed to get from Lori s house to Alex s house is the difference between the y-coordinate of Alex s House (, 4) Lori s house, 3, and the y-coordinate of Alex s house, 4, or 3 ( 4) = = 7. From Lori s house, Alex s house is left and down. Therefore the translation needed to get from Lori s house to Alex s house is 8 units left and 7 units down. y Lori s House (6, 3) x Answer: (D) 16. Reading from the graph, Riley-Ann scored 8 points in Game 1, 7 points in Game, 0 points in Game 3, 7 points in Game 4, and 18 points in Game 5. Therefore the mean number of points that she scored per game is = 60 = Since the ordered list (smallest to largest) of the number of points scored per game is 7, 7, 8, 18, 0, then the median is 8, the number in the middle of this ordered list. The difference between the mean and the median of the number of points that Riley-Ann scored is 1 8 = 4. Answer: (D)

23 014 Gauss Contest s Page Since P QR is a straight line segment, then P QR = 180. Since SQP + SQR = 180, then SQR = 180 SQP = = 105. The three angles in a triangle add to 180, so QSR + SQR + QRS = 180, or QSR = 180 SQR QRS = = 45. The exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles of the triangle. Since SQP is an exterior angle of SQR, and the two opposite interior angles are QSR and QRS, then SQP = QSR + QRS. Thus, 75 = QSR + 30 or QSR = = 45. Answer: (E) The outer square has an area of 9 cm, so the sides of this outer square have length 3 cm (since 3 3 = 9), and thus P N = 3 cm. The inner square has an area of 1 cm, so the sides of this inner P Q M square have length 1 cm (since 1 1 = 1), and thus MR = 1 cm. Since P N = 3 cm, then P S + SN = 3 cm and so QR + SN = 3 cm (since QR = P S). But QR = QM + MR, so then QM + MR + SN = 3 cm or QM SN = 3 cm (since MR = 1 cm). From this last equation we get QM + SN = cm. S N R Since each of QM and SN is the width of an identical rectangle, then QM = SN = 1 cm. Using P S + SN = 3 cm, we get P S + 1 = 3 cm and so P S = cm. Since the rectangles are identical, then SN = P Q = 1 cm. The perimeter of rectangle P QRS is (P S + P Q) = ( + 1) = 3 = 6 cm. The outer square has an area of 9 cm, so the sides of this outer square have length 3 cm (since 3 3 = 9), and thus P N = 3 cm. Since P N = 3 cm, then P S + SN = 3 cm. Since each of P Q and SN is the width of an identical rectangle, then P Q = SN and so P S + SN = P S + P Q = 3 cm. The perimeter of P QRS is (P S + P Q) = 3 = 6 cm. N Answer: (A) 19. The width of Sarah s rectangular floor is 18 hand lengths. Since each hand length is 0 cm, then the width of the floor is 18 0 = 360 cm. The length of Sarah s rectangular floor is hand lengths. Since each hand length is 0 cm, then the length of the floor is 0 = 440 cm. Thus the area of the floor is = cm. Of the given answers, the closest to the area of the floor is cm. Answer: (A) 0. 1 Since = 8000 and = 7 000, then we might guess that the three consecutive odd numbers whose product is 9177 are closer to 0 than they are to 30. Using trial and error, we determine that = 1 075, which is too large. P S Q M R

24 014 Gauss Contest s Page 7 The next smallest set of three consecutive odd numbers is 19, 1, 3 and the product of these three numbers is = 9177, as required. Thus, the sum of the three consecutive odd numbers whose product is 9177 is = 63. We begin by determining the prime numbers whose product is (This is called the prime factorization of 9177.) This prime factorization of 9177 is shown in the factor tree to the right. That is, 9177 = = = Since 3 7 = 1, then 9177 = and so the three consecutive numbers whose product is 9177 are 19, 1, 3. Thus, the sum of the three consecutive odd numbers whose product is 9177 is = Answer: (D) 1. At Store Q, the bicycle s regular price is 15% more than the price at Store P, or 15% more than $00. Since 15% of 00 is = = 30, then 15% more than $00 is $00+$30 or $ This bicycle is on sale at Store Q for 10% off of the regular price, $30. Since 10% of 30 is = = 3, then 10% off of $30 is $30 $3 or $ The sale price of the bicycle at Store Q is $07. Answer: (D). Assume the top face of the cube is coloured green. Since the front face of the cube shares an edge with the top face, it cannot be coloured green. Thus, we need at least two colours. Thus, we assume that the front face is coloured blue, as shown in Figure 1. Since the right face shares an edge with the top face and with the front face, it cannot be coloured green or blue. Thus, we need at least three colours. Thus, we assume that the right face is coloured red, as shown in Figure. We have shown that at least 3 colours are needed. In fact, the cube can be coloured with exactly 3 colours by colouring the left face red, the back face blue, and the bottom face green (Figure 3). In this way, the cube is coloured with exactly 3 colours and no two faces that share an edge are the same colour. Therefore, 3 is the smallest number of colours needed to paint a cube so that no two faces that share an edge are the same colour. blue Figure 1 Figure Figure 3 green green green blue blue red red blue red green Answer: (B)

25 014 Gauss Contest s Page For each of the 6 possible outcomes that could appear on the red die, there are 6 possible outcomes that could appear on the blue die. That is, the total number of possible outcomes when a standard six-sided red die and a standard six-sided blue die are rolled is 6 6 = 36. These 36 outcomes are shown in the table below. When a number that appears on the red die is greater than a number that appears on the blue die, a checkmark has been placed in the appropriate cell, corresponding to the intersection of the column and row. For example, the table cell containing the double checkmark represents the outcome of a 4 appearing on the red die and a appearing on the blue die. Number on the Blue Die Blue Red Number on the Red Die Of the 36 possible outcomes, or 15 have a number appearing on the red die that is larger than the number appearing on the blue die. The probability that the number appearing on the red die is greater than the number appearing on the blue die is As in 1, we determine the total number of possible outcomes to be 36. Each of these 36 outcomes can be grouped into one of three possibilities; the number appearing on the red die is greater than the number appearing on the blue die, the number appearing on the red die is less than the number appearing on the blue die, or the numbers appearing on the two dice are equal. There are 6 possible outcomes in which the numbers appearing on the two dice are equal (both numbers are 1, both numbers are, and so on). Of the 36 total outcomes, this leaves 36 6 = 30 outcomes in which either the number appearing on the red die is greater than the number appearing on the blue die, or the number appearing on the red die is less than the number appearing on the blue die. These two possibilities are equally likely to happen (since both dice are identical except for colour), and so the number appearing on the red die will be greater than the number appearing on the blue die in half of the 30 outcomes, or 15 outcomes. Thus, the probability that the number appearing on the red die is greater than the number appearing on the blue die is Answer: (C)

26 014 Gauss Contest s Page 9 4. We begin by joining Q to P. Since Q and P are the midpoints of ST and UV, then QP is parallel to both SV and T U and rectangles SQP V and QT UP are identical. In rectangle SQP V, V Q is a diagonal. Similarly, since P R is parallel to V Q then P R extended to T is a diagonal of rectangle QT UP, as shown in Figure 1. In Figure, we label points A, B, C, D, E, and F, the midpoints Figure 1 of SQ, QT, T U, UP, P V, and V S, respectively. We join A to E, B to D and F to C, with F C intersecting QP S A Q B T at the centre of the square O, as shown. Since P R = QR and R lies on diagonal P T, then both F C and BD pass through R. (That is, R is the centre of QT UP.) F C The line segments AE, QP, BD, and F C divide square ST UV O R into 8 identical rectangles. In one of these rectangles, QBRO, diagonal QR divides the V E P D U rectangle into equal areas. Figure That is, the area of QOR is half of the area of rectangle QBRO. Similarly, the area of P OR is half of the area of rectangle P ORD. Rectangle SQP V has area equal to 4 of the 8 identical rectangles. Therefore, QP V has area equal to of the 8 identical rectangles (since diagonal V Q divides the area of SQP V in half). Thus the total shaded area, which is QOR + P OR + QP V, is equivalent to the area of or 3 of the identical rectangles. Since square ST UV is divided into 8 of these identical rectangles, and the shaded area is equivalent to the area of 3 of these 8 rectangles, then the unshaded area occupies an area equal to that of the remaining 8 3 or 5 rectangles. Therefore, the ratio of the shaded area to the unshaded area is 3 : 5. Answer: (B) 5. We begin by listing the first few line segment endpoints as determined by the number of segments that Paul has drawn. Line Segment Number Endpoint of the Line Segment 1 (1, 0) (1, ) 3 (4, ) 4 (4, 6) 5 (9, 6) 6 (9, 1) 7 (16, 1) 8 (16, 0) Since each new line segment is drawn either vertically or horizontally from the endpoint of the previous line segment, then only one of the x-coordinate or y-coordinate changes from one endpoint to the next endpoint. Further, since the odd numbered line segments are horizontal, the x-coordinate of the endpoint of these segments changes from the previous endpoint while the y-coordinate remains the same. S V Q P R T U

27 014 Gauss Contest s Page 10 Similarly, since the even numbered line segments are vertical, the y-coordinate of the endpoint of these segments changes from the previous endpoint while the x-coordinate remains the same. We are given that one of the line segments ends at the point (59, 506). We will begin by determining the number of segments that must be drawn for the x-coordinate of the endpoint of the final line segment to be 59. The first line segment is drawn horizontally from the origin, to the right, has length 1, and thus ends with an x-coordinate of 1. The third line segment is drawn horizontally to the right, has length 3 and thus ends with an x-coordinate of = 4. The fifth line segment is drawn horizontally to the right, has length 5 and thus ends with an x-coordinate of = 9. To get an idea of how many line segments are required before the x-coordinate of the endpoint is 59, we begin by considering the first 1 line segments. The endpoint of the 1st line segment has x-coordinate equal to To make this sum easier to determine, we rearrange the terms to get (1+1)+(3+19)+(5+17)+(7+15)+(9+13)+11 = = 5+11 = 11. Since 11 is much smaller than 59, we continue to try larger numbers of line segments until we finally reach 45 line segments. The endpoint of the 45th line segment has x-coordinate equal to To make this sum easier to determine, we rearrange the terms to get (1 + 45) + (3 + 43) + (5 + 41) + + (19 + 7) + (1 + 5) + 3 = = 59. Therefore, the 45th line segment ends with an x-coordinate of 59. The next line segment (the 46th) will also have an endpoint with this same x-coordinate, 59, since even numbered line segments are vertical (thus, only the y-coordinate will change). This tells us that (59, 506) is the endpoint of either the 45th or the 46th line segment. At this point we may confirm that 506 is the y-coordinate of the endpoint of the 45th line segment (and hence the 44th line segment as well). The second line segment is drawn vertically from the x-axis, upward, has length, and thus ends with y-coordinate. The fourth line segment is drawn vertically upward, has length 4 and thus ends with a y-coordinate of + 4 = 6. The sixth line segment is drawn vertically upward, has length 6 and thus ends with a y-coordinate of = 1. The endpoint of the 45th line segment has y-coordinate equal to Rearranging the terms of this sum, we get ( + 44) + (4 + 4) + (6 + 40) + + (0 + 6) + ( + 4) = = 506. Thus, (59, 506) is the endpoint of the 45th line segment. The 46th line segment is vertical and has length 46. Therefore the y-coordinate of the endpoint of the 46th line segment is = = 55. Since the x-coordinate of the endpoint of the 46th line segment is the same as that of the 45th, then the endpoint of the next line segment that Paul draws is (59, 55). Answer: (A)

28 014 Gauss Contest s Page 11 Grade 8 1. The number is ten thousand one hundred one. Therefore, it is equal to Since one scoop can feed 8 goldfish, then 4 scoops can feed 4 8 = 3 goldfish. 3. Evaluating, ( ) (013 01) = 1 1 = 1. Answer: (D) Answer: (E) Answer: (B) 4. The measure of the three angles in any triangle add to 180. Since two of the angles measure 90 and 55, then the third angle in the triangle measures = 35. The measure of the smallest angle in the triangle is 35. Answer: (D) 5. The positive value of any number is equal to the distance that the number is away from zero along a number line. The smaller the distance a number is away from zero along a number line, the closer the number is to zero. Ignoring negative signs (that is, we consider all 5 answers to be positive), the number that is closest to zero is the smallest number. The smallest number from the list {1101, 1011, 1010, 1001, 1110} is Of the given answers, the number 1001 is the closest number to zero. Answer: (D) 6. Since 5y 100 = 15, then 5y = = 5, and so y = 5 5 = 45. Answer: (A) 7. Each prime number is divisible by exactly two numbers, 1 and itself. Each of the even numbers {1, 14, 16, 18, 0,, 4, 6, 8} is divisible by 1 and itself and also by. Therefore the numbers {1, 14, 16, 18, 0,, 4, 6, 8} are not prime. Each of the numbers {15, 1, 7} is divisible by 1 and itself and also by 3. Therefore the numbers {15, 1, 7} are not prime. The number 5 is divisible by 1 and itself and also by 5, so it is not prime. Each of the remaining numbers is divisible by exactly 1 and itself. Thus, the list of prime numbers between 10 and 30 is {11, 13, 17, 19, 3, 9}. There are 6 prime numbers between 10 and 30. Answer: (C) 8. The given triangle is isosceles, so the unmarked side also has length x cm. The perimeter of the triangle is 53 cm, so x + x + 11 = 53 or x + 11 = 53 or x = = 4, and so x = 4 = 1. Answer: (B)

29 014 Gauss Contest s Page To order the fractions { 3, 3, 6, } from smallest to largest, we first express each fraction with a common denominator of 7 5 = 70. The set { 3, 3, 6, } { is equivalent to the set 3 10, 3 35, 6 10, } { or to the set 30, 105, 60, }. Ordered from smallest to largest, the set is { 30, 4, 60, } , so ordered from smallest to largest the original set is { 3, 3, 6, }. Since the numerators of the given fractions are each either 3 or 6, we rewrite each fraction with a common numerator of 6. The set { 3, 3, 6, } { is equivalent to the set 3, 3, 6, } { or to the set 6, 6, 6, }. Since the numerators are equal, then the larger the denominator, the smaller the fraction. Ordered from smallest to largest, the set is { 6, 6, 6, }, so ordered from smallest to largest the original set is { 3, 3, 6, }. Answer: (A) The ratio of the number of girls to the number of boys is 3 : 5, so for every 3 girls there are 5 boys and = 8 students. That is, the number of girls in the class is 3 of the number of students in the class. 8 Since the number of students in the class is 4, the number of girls is 3 4 = 7 = The remaining 4 9 = 15 students in the class must be boys. Therefore, there are 15 9 = 6 fewer girls than boys in the class. The ratio of the number of girls to the number of boys is 3 : 5, so if there are 3x girls in the class then there are 5x boys. In total, there are 4 students in the class, so 3x + 5x = 4 or 8x = 4 and so x = 3. The number of girls in the class is 3x = 3(3) = 9 and the number of boys in the class is 5x = 5(3) = 15. Therefore, there are 15 9 = 6 fewer girls than boys in the class. 3 The ratio of the number of girls to the number of boys is 3 : 5, so for every 3 girls there are 5 boys and = 8 students. That is, 3 of the number of students in the class are girls and 5 of the number of students are 8 8 boys. Thus, the difference between the number of boys in the class and the number of girls in the class is 5 3 = = 1 of the number of students in the class Therefore, there are 1 4 = 6 fewer girls than boys in the class. 4 Answer: (D) 11. Since there are 7 days in a week, any multiple of 7 days after a Wednesday is also a Wednesday. Since 70 is a multiple of 7, then 70 days after John was born, the day is also a Wednesday. Therefore, 71 days after John was born is a Thursday and 7 days after John was born is a Friday. Alison was born on a Friday. Answer: (E)

30 014 Gauss Contest s Page Opposite angles have equal measures. Since AOC and DOB are opposite angles, then y = 40. Straight angle COD measures 180. Since AOC + AOD = COD, then 40 + x = 180 and so x = 140. The difference x y is = 100. A C x o y o 40 o O D B Answer: (E) Since the scores in each of the 5 sets are listed in order, from smallest to largest, the number in the middle (the 3rd number) of the 5 numbers is the median. The mean is calculated and also shown below for each of the 5 sets of scores. Set of Scores Median Mean 10, 0, 40, 40, = = 30 40, 50, 60, 70, , 0, 0, 50, = = 60 = = 38 10, 0, 30, 100, , 50, 50, 50, = = 7 = = 60 From the table above, we see that the first set of scores is the only set in which the median, 40, is greater than the mean, 30. The first set of 5 scores, 10, 0, 40, 40, 40, is ordered from smallest to largest and thus the median is the 3rd (middle) number of this set, 40. The mean of the final 3 numbers of this set is also 40, since each is equal to 40. Since the first numbers of the set are both less than the mean of the other 3 numbers (40), then the mean of all 5 numbers in the set must be less than 40. Using a similar argument for the remaining four answers, we see that each of the other four sets has its mean equal to or greater than its median. Therefore, the first set of scores is the only set in which the median is greater than the mean. Answer: (A) 14. Betty has 3 equally likely choices for the flavour of ice cream (chocolate or vanilla or strawberry). For each of these 3 choices, she has equally likely choices for the syrup (butterscotch or fudge). Thus, Betty has 3 or 6 equally likely choices for the flavour and syrup of the sundae. For each of these 6 choices, Betty has 3 equally likely choices for the topping (cherry or banana or pineapple). This makes 6 3 or 18 equally likely choices for her sundae. Since only 1 of these 18 choices is the sundae with vanilla ice cream, fudge syrup and banana topping, then the probability that Betty randomly chooses this sundae is Answer: (A)

31 014 Gauss Contest s Page After reflecting the point A(1, ) in the y-axis, the y-coordinate of the image will be the same as the y-coordinate of point A, y =. Point A is a distance of 1 to the right of the y-axis. The image will be the same distance from the y-axis, but to the left of the y-axis. Thus, the image has x-coordinate 1. The coordinates of the reflected point are ( 1, ). (-1, ) -1 y 1 1 A(1, ) x Answer: (B) Construct P H perpendicular to AB, as shown. The area of ABP is 40 and so 1 D P C AB P H = 40, or AB P H = 80, and since AB = 10, then P H = 8. Since CB = P H = 10, the area of ABCD is 10 8 = 80. The shaded area equals the area of ABP subtracted from the area of ABCD, or = 40. A H B 10 As in 1, we construct P H perpendicular to AB. Since both DA and CB are perpendicular to AB, then P H is parallel to DA and to CB. That is, DAHP and P HBC are rectangles. Diagonal P A divides the area of rectangle DAHP into two equal areas, P AH and P AD. Diagonal P B divides the area of rectangle P HBC into two equal areas, P BH and P BC. Therefore, the area of P AH added to the area of P BH is equal to the area of P AD added to the area of P BC. However, the area of P AH added to the area of P BH is equal to the area of ABP, which is 40. So the area of P AD added to the area of P BC is also 40. Therefore, the area of the shaded region is 40. Answer: (B) 17. Of the 10 multiple choice questions, Janine got 80% or = 8 correct. Of the 30 short answer questions, Janine got 70% or = 1 correct. In total, Janine answered 8+1 = 9 of the 40 questions correct, or 9 100% = 7.5% correct. 40 Answer: (B) 18. The area of the rectangle, 48 cm, is given by the product of the rectangle s length and width, and so we must consider all possible pairs of whole numbers whose product is 48. In the table below, we systematically examine all possible factors of 48 in order to determine the possible whole number side lengths of the rectangle and its perimeter. Factors of 48 Side Lengths of Rectangle Perimeter of Rectangle 48 = and 48 (1 + 48) = 49 = = 4 and 4 ( + 4) = 6 = 5 48 = and 16 (3 + 16) = 19 = = and 1 (4 + 1) = 16 = 3 48 = and 8 (6 + 8) = 14 = 8 The side lengths of the rectangle having area 48 cm and perimeter 3 cm, are 4 and 1. In cm, the positive difference between the length and width of the rectangle is 1 4 = 8. Answer: (D)

32 014 Gauss Contest s Page At Store Q, the bicycle s regular price is 15% more than the price at Store P, or 15% more than $00. Since 15% of 00 is = = 30, then 15% more than $00 is $00+$30 or $ This bicycle is on sale at Store Q for 10% off of the regular price, $30. Since 10% of 30 is = = 3, then 10% off of $30 is $30 $3 or $ The sale price of the bicycle at Store Q is $07. Answer: (D) 0. Using 7 of the 5 stamps (35 ) and 1 of the 8 stamps (8 ) makes 43 in postage. Also, 3 of the 5 stamps (15 ) and 3 of the 8 stamps (4 ) makes 39 in postage. This eliminates the choices (D) and (E). Of the five given answers, the next largest is 7. To make 7 in postage, 0, 1,, or 3 of the 8 stamps must be used (using more than 3 of the 8 stamps would exceed 7 in postage). If 0 of the 8 stamps are used, then the remaining 7 must be made from 5 stamps. This is not possible since 7 is not a multiple of 5. If 1 of the 8 stamps are used, then the remaining 7 8 = 19 must be made from 5 stamps. This is not possible since 19 is not a multiple of 5. If of the 8 stamps are used, then the remaining 7 16 = 11 must be made from 5 stamps. This is not possible since 11 is not a multiple of 5. If 3 of the 8 stamps are used, then the remaining 7 4 = 3 must be made from 5 stamps. This is clearly not possible. Therefore of the five given answers, 7 is the largest amount of postage that cannot be made using only 5 and 8 stamps. It is worth noting that all amounts of postage larger than 7 can be made using only 5 and 8 stamps. To see why this is true, first consider that all five postage amounts from 8 to 3 can be made as shown in the table below. Number of 5 Stamps Number of 8 Stamps Value of Stamps 4 1 (4 5 )+(1 8 )= (1 5 )+(3 8 )= (6 5 )+(0 8 )= 30 3 (3 5 )+( 8 )= (0 5 )+(4 8 )= 3 The next five amounts of postage, 33 to 37, can be made by adding 1 additional 5 stamp to each of the previous five amounts of postage. That is, = 33, and = 34, and so on. We can make every postage amount larger than 7 by continuing in this way to add 1 additional 5 stamp to each of the five amounts from the previous group. Answer: (C) 1. The shaded area in the top three rows of the diagram contains 6 circles with radius 1 cm. The shaded area in the bottom row of the diagram contains 4 semi-circles with radius 1 cm, whose combined area is equal to the area of circles with radius 1 cm. In total, the shaded area of the diagram contains 6 + = 8 circles with radius 1 cm. In cm, the total shaded area is 8 π 1 = 8π. Answer: (E)

33 014 Gauss Contest s Page 16. We first determine the surface area of the original cube, before the two cubes were cut from its corners. The original cube had 6 identical square faces, each of whose area was 3 3 = 9 cm. Thus, the surface area of the original cube was 6 9 = 54 cm. Next we will explain why the resulting solid in question has surface area equal to that of the original cube, 54 cm. Consider the front, bottom right corner of the resulting solid, as shown and labeled in Figure 1. Cutting out a 1 cm by 1 cm by 1 cm cube from this corner exposes 3 new square faces, RSP Q, RST U and SP W T, which were not a part of the surface area of the original cube. Next we consider what surface area was part of the original cube that is not present in the resulting solid. To visualize this, we replace the vertex V of the original cube and reconstruct segments QV, UV and W V, as shown in Figure. We note that P QRST UV W is the 1 cm by 1 cm by 1 cm cube that was cut from the corner of the original cube. In Figure, we see that the surface areas that were part of the original cube that are not present in the resulting solid are the 3 square faces UT W V, QP W V and RQV U. R U R U S S T T Figure 1 Figure To summarize, removal of the 1 cm by 1 cm by 1 cm cube from the corner of the original cube exposes 3 new square faces, RSP Q, RST U and SP W T, which were not a part of the surface area of the original cube. These 3 faces represent half of the surface area of cube P QRST UV W (since they are 3 of the 6 identical faces of the cube). However, in removing the 1 cm by 1 cm by 1 cm cube from the corner of the original cube, the 3 faces, UT W V, QP W V and RQV U, are lost. These 3 faces represent the other half of the surface area of cube P QRST UV W. That is, the total surface area gained by removal of the 1 cm by 1 cm by 1 cm cube is equal to the total surface area lost by removing this same cube. This same argument can be made for the removal of the cm by cm by cm cube from the top, back left corner of the original cube. Thus the surface area of the resulting solid is equal to the surface area of the original cube, which was 54 cm. Answer: (E) 3. The first positive odd integer is 1, the second is () 1 = 3, the third is (3) 1 = 5, the fourth is (4) 1 = 7. That is, the one hundredth positive odd integer is (100) 1 = 199 and the sum that we are being asked to determine is Since each odd integer can be expressed as the sum of two consecutive integers, we rewrite as 1 + (1 + ) + ( + 3) + + ( ) + ( ) + ( ). Rearranging the terms in the previous sum, we get ( ) + ( ). Q V Q P W P W

34 014 Gauss Contest s Page 17 The sum in the first set of brackets, is equal to The sum in the second set of brackets, is 100 less than 5050 or Therefore, = = Doubling each term on the left of the equality = 5050, doubles the result on the right. That is, = Subtracting 1 from each term on the left side of this equality gives ( 1)+(4 1)+(6 1)+ +(196 1)+(198 1)+(00 1) or , the required sum. Since there are 100 terms on the left side of the equality, then subtracting one 100 times reduces the left side of the equality by 100. Subtracting 100 from the right side, we get = , and so = It is also possible to determine the sum of the first 100 positive odd integers, , without using the result given in the question. Reorganizing the terms into sums of pairs, first the largest number with the smallest, then the next largest number with the next smallest, and so on, = (1+199)+(3+197)+(5+195)+ +(97+103)+(99+101). The sum of the first pair, is 00. Each successive pair following includes a number that is more than a number in the previous pair, and also a number that is less than a number in the previous pair, so then the sum of every pair is also 00. The total number of terms in the sum is 100, and thus there are 50 pairs of numbers each of which has a sum of 00. Therefore, = = Answer: (B) 4. The original 4 4 grid contains exactly 30 squares, of which 16 are of size 1 1, 9 are of size, 4 are of size 3 3, and 1 is of size 4 4 ( = 30). In each of the 5 answers given, exactly one or two of the 1 1 squares are missing from the original grid. We determine the number of these 30 squares that cannot be formed as a result of these missing 1 1 squares, and subtract this total from 30 to find the total number of squares in each of the 5 given configurations. Number of Missing Squares Total Number of Squares A = 6 B = 4 C = 3 D = 3 E = 3

35 014 Gauss Contest s Page 18 For clarity, the 6 squares that exist in the original grid but that are missing from the grid in answer B are shown below by size. 1x1 x x 3x3 3x3 4x4 The grid given in answer B contains exactly 4 squares. Answer: (B) 5. A total of N people participated in the survey. Exactly 9 9 of those surveyed, or N people, said that the colour of the flower was important Exactly 7 7 of those surveyed, or N people, said that the smell of the flower was important. 1 1 Since the number of people surveyed must be a whole number, then N must be a multiple of 14 and N must also be a multiple of 1. The lowest common multiple of 14 and 1 is 7 6 or 84, and so N is a multiple of 84. Since N is a multiple of 84, let N = 84k where k is some positive integer. 9 That is, N = 9 (84k) = 54k people said that the colour of the flower was important and N = 7 (84k) = 49k people said that the smell of the flower was important. 1 1 The information can be represented with a Venn diagram, as shown below. People Surveyed (84k) Colour (54k) Smell (49k) 54k k Since N = 84k, we must find both a maximum and a minimum value for k in order to determine the number of possible values for N. Finding a Minimum Value for k We know that 753 people said that both the colour and the smell were important. We also know that 54k said that the colour was important and that 49k said that the smell was important. The 753 people who said that both were important are included among both the 54k people and among the 49k people. Therefore, 54k must be at least 753 and 49k must also be at least 753. Since 54k is larger than 49k, then we only need to find k for which 49k is larger than or equal to 753. We note that = 735 and that = 784. Therefore, the smallest value of k for which 49k is at least 753 is k = 16. As noted above, k = 16 will also make 54k larger than 753. Therefore, the minimum possible value for k is 16.

36 014 Gauss Contest s Page 19 Finding a Maximum Value for k There were 84k people surveyed of whom 54k said that the colour was important. This means that there are 84k 54k = 30k people who said that the colour was not important. We note that 49k people said that the smell was important. These 49k people include some or all of the 30k people who said that the colour was not important. In other words, at most 30k of the 49k people said that the smell was important and the colour was not important, which means that at least 49k 30k = 19k people said that the smell was important and that the colour was important. Since we know that 753 people said that both the smell and colour were important, then 753 is at least as large as 19k. Since = 760 and = 741, then k must be at most 39. The minimum value of k is 16 and the maximum value of k is 39. Thus, there are = 4 possible values for k and since N = 84k, there are also 4 possible values for N. Note: We should also justify that each value of k between 16 and 39 is possible. Using the Venn diagram above, we see that the total of 84k people surveyed means that 84k (54k 753) 753 (49k 753) = k people fall outside the two circles. From our work in finding the minimum above, we can see that for each k between 16 and 39, inclusive, each of the three integers 54k 753, 753, and 49k 753 is positive, since k is at least 16. Also, from our work in finding the maximum above, we can see that for each k between 16 and 39, inclusive, the integer k is positive, since k is at most 39. Therefore, all four quantities are positive integers for each of these values of k, which means that we can construct a Venn diagram with these numbers, as required. Answer: (D)

37 The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 013 Gauss Contests (Grades 7 and 8) Wednesday, May 15, 013 (in North America and South America) Thursday, May 16, 013 (outside of North America and South America) s 01 University of Waterloo

38 013 Gauss Contest s Page 3 Grade 7 1. Evaluating, (5 3) = 15 = 13. Answer: (E). 1 A number is a multiple of 9 if it is the result of multiplying 9 by an integer. Of the answers given, only 45 results from multiplying 9 by an integer, since 45 = 9 5. A number is a multiple of 9 if the result after dividing it by 9 is an integer. Of the answers given, only 45 results in an integer after dividing by 9, since 45 9 = 5. Answer: (D) 3. Thirty-six hundredths equals or Answer: (A) 4. By grouping terms using brackets as shown, (1 + 1 ) + ( ) + ( ), we can see that the result inside each set of brackets is 0. Thus the value of is 0. Answer: (D) 5. Since P Q is a straight line segment, the three angles given sum to 180. That is, 90 + x + 0 = 180 or x = or x = 70. Answer: (B) 6. Nick has 6 nickels and each nickel is worth 5. So Nick has 6 5 or 30 in nickels. Nick has dimes and each dime is worth 10. So Nick has 10 or 0 in dimes. Nick has 1 quarter and each quarter is worth 5. So Nick has 1 5 or 5 in quarters. In total, Nick has or 75. Answer: (B) 7. 1 To determine the smallest number in the set { 1,, 1, 5, }, we express each number with a common denominator of 1. The set { 1,, 1, 5, or to the set { 6, 8, 3, 10, }. The smallest number in this set is 3 1, so 1 4 } is equivalent to the set { 1 6 6, 4 3 4, , 5 6, 7 1 is the smallest number in the original set. With the exception of 1, each number in the set is greater than or equal to 1. 4 We can see this by recognizing that the numerator of each fraction is greater than or equal to one half of its denominator. Thus, 1 is the only number in the list that is less than 1 and so it must be the smallest number 4 in the set. Answer: (C) 8. Since Ahmed stops to talk with Kee one quarter of the way to the store, then the remaining distance to the store is 1 1 = 3 of the total distance. 4 4 Since 3 = 3 1, then the distance that Ahmed travelled from Kee to the store is 3 times the 4 4 distance that Ahmed travelled from the start to reach Kee. That is, 1 km is 3 times the distance between the start and Kee. So the distance between the start and Kee is 1 = 4 km. Therefore, the total distance travelled by Ahmed is 4+1 or 16 km. 3 Answer: (B) }

39 013 Gauss Contest s Page 4 9. When n = 4, we are looking for an expression that produces a value of 7. The results of substituting n = 4 into each expression and evaluating are shown below. Expression Value (A) 3n 3(4) = 1 = 10 (B) (n 1) (4 1) = (3) = 6 (C) n = 8 (D) n (4) = 8 (E) n 1 (4) 1 = 8 1 = 7 Since n 1 is the only expression which gives a value of 7 when n = 4, it is the only possible answer. We check that the expression n 1 does give the remaining values, 1, 3, 5, 9, when n = 1,, 3, 5, respectively. (Alternately, we may have began by substituting n = 1 and noticing that this eliminates answers (B), (C) and (D). Substituting n = eliminates (A). Substituting n = 3, n = 4 and n = 5 confirms the answer (E).) Answer: (E) 10. To make the difference UV W XY Z as large as possible, we make UV W as large as possible and XY Z as small as possible. The hundreds digit of a number contributes more to its value than its tens digit, and its tens digit contributes more to its value than its units digit. Thus, we construct the largest possible number UV W by choosing 9 (the largest digit) to be its hundreds digit, U, and by choosing 8 (the second largest digit) to be its tens digit, V, and by choosing 7 (the third largest digit) to be the units digit, W. Similarly, we construct the smallest possible number XY Z by choosing 1 (the smallest allowable digit) to be its hundreds digit, X, and (the second smallest allowable digit) to be its tens digit, Y, and by choosing 3 (the third smallest allowable digit) to be its units digit, Z. The largest possible difference is UV W XY Z or or 864. Answer: (B) 11. Each face of a cube is a square. The dimensions of each face of the cube are 1 cm by 1 cm. Thus, the area of each face of the cube is 1 1 = 1 cm. Since a cube has 6 identical faces, the surface area of the cube is 6 1 = 6 cm. Answer: (E) 1. The greatest common factor of two numbers is the largest positive integer which divides into both numbers with no remainder. For answer (B), 50 divided by 0 leaves a remainder, so we may eliminate (B) as a possible answer. Similarly for answer (D), 5 divided by 0 leaves a remainder, so we may eliminate (D) as a possible answer. For answer (A), since 00 divides 00 and 00 divides 000, the greatest common factor of 00 and 000 cannot be 0. For answer (E), since 40 divides 40 and 40 divides 80, the greatest common factor of 40 and 80 cannot be 0. The largest positive integer which divides both 0 and 40 is 0, and so (C) is the correct answer. Answer: (C)

40 013 Gauss Contest s Page Since Lan and Mihai are seated beside each other, while Jack and Kelly are not, the only possible location for the remaining chair (Nate s chair) is between Jack and Kelly, as shown. Therefore, the people who are seated on either side of Nate are Jack and Kelly. Jack Lan Mihai Kelly Answer: (B) 14. Substituting x = 4 into 3x + y we get, 3(4) + y = 1 + y. Since this expression 1 + y is equal to 30, then y must equal 30 1 or 18. If y = 18, then y is 18 or 9. Answer: (E) 15. Each time Daniel reaches into the jar, he removes half of the coins that are in the jar. Since he removes half of the coins, then the other half of the coins remain in the jar. We summarize Daniel s progress in the table below. Number of Times Coins are Removed Number of Coins Remaining in the Jar For exactly 1 coin to remain in the jar, Daniel must reach in and remove coins from the jar 6 times. Answer: (C) Consider the set of five consecutive even numbers 8, 10, 1, 14, 16. The mean of these five numbers is = 60 or If the five consecutive even numbers were smaller, their mean would be less than 1. If the five consecutive even numbers were larger, their mean would be greater than 1. Therefore, this is the set of five consecutive even numbers that we seek. The mean of the smallest and largest of these five numbers is 8+16 = 4 or 1. The mean of five consecutive even numbers is the middle (third largest) number. To see this, consider that the smallest of the five numbers is 4 less than the middle number, while the largest of the five numbers is 4 more than the middle number. Thus, the mean of the smallest and largest numbers is the middle number. Similarly, the second smallest of the five numbers is less than the middle number, while the fourth largest of the five numbers is more than the middle number. Thus, the mean of these two numbers is also the middle number. Since the mean of the five numbers is 1, then the middle number is 1. Therefore, the mean of the largest and smallest of the five numbers is also 1. Answer: (A) 17. For every 3 chocolates that Claire buys at the regular price, she buys a fourth for 5 cents. Consider dividing the 1 chocolates that Claire buys into 3 groups of 4 chocolates. In each group of 4, Claire buys 3 chocolates at the regular price and the fourth chocolate is

41 013 Gauss Contest s Page 6 purchased for 5 cents. That is, of the 1 chocolates that Claire buys, 3 are bought at 5 cents each while the remaining 9 are purchased at the regular price. The total cost to purchase 3 chocolates at 5 cents each is 3 5 = 75 cents. Since Claire spent $6.15 and the total cost of the 5 cent chocolates was 75 cents, then the cost of the regular price chocolates was $6.15 $0.75 = $5.40. Since 9 chocolates were purchased at the regular price for a total of $5.40, then the regular price of one chocolate is $5.40 = $0.60 or 60 cents. 9 Answer: (C) 18. The total area of the shaded regions is the difference between the area of square JKLM and the area of the portion of rectangle P QRS that overlaps JKLM. Since JK = 8, then the area of square JKLM is 8 8 or 64. Since JK is parallel to P Q, then the portion of P QRS that overlaps JKLM is a rectangle, and has length equal to JK or 8, and width equal to P S or. So the area of the portion of P QRS that overlaps JKLM is 8 = 16. Therefore the total area of the shaded regions is = 48. Answer: (D) 19. Using the special six-sided die, the probability of rolling a number that is a multiple of three is 1. Since 1 of 6 is 3, then exactly 3 numbers on the die must be multiples of 3. Since the probability of rolling an even number is 1 and 1 of 6 is, then exactly numbers on 3 3 the die must be even. The die in (A) has only numbers that are multiples of 3 (3 and 6), and thus may be eliminated. The die in (C) has 4 numbers that are even (, 4, 6, 6), and thus may be eliminated. The die in (D) has 3 numbers that are even (, 4, 6), and thus may be eliminated. The die in (E) has 4 numbers that are multiples of 3 (3, 3, 3, 6), and thus may be eliminated. The die in (B) has exactly 3 numbers that are multiples of 3 (3, 3, 6), and exactly even numbers ( and 6), and is therefore the correct answer. Answer: (B) 0. In the diagram shown, the 31 identical toothpicks used in the 1 10 grid are separated into sections. The top section, T, is made from 11 vertical toothpicks and 10 horizontal toothpicks, or 1 toothpicks in total. The bottom section, B, is made of 10 horizontal toothpicks. The 10 and 3 10 grids are similarly separated into top and bottom sections, as shown. We observe that the 1 10 grid consists of 1 top section and 1 bottom. The 10 grid consists of top sections and 1 bottom. The 3 10 grid consists of 3 top sections and 1 bottom. Continuing in this way, a grid of size n 10 will consist of n top sections and 1 bottom section, for any positive integer n. So then a grid of size consists of 43 top sections and 1 bottom section. Each top section is made from 1 toothpicks and each bottom section is made from 10 toothpicks. Thus, the total number of toothpicks in a grid is (43 1) + (1 10) or or T B T T B T T T B

42 013 Gauss Contest s Page 7 1. The sum of the units column is P + P + P = 3P. Since P is a single digit, and 3P ends in a 7, then the only possibility is P = 9. This gives: Q 9 + Q Q Answer: (A) Then 3P = 3 9 = 7, and thus is carried to the tens column. The sum of the tens column becomes Q + Q or 9 + Q. Since 9 + Q ends in a 9 (since P = 9), then Q ends in 9 9 = 0. Since Q is a single digit, there are two possibilities for Q such that Q ends in 0. These are Q = 0 and Q = 5. If Q = 0, then the sum of the tens column is 9 with no carry to the hundreds column. In this case, the sum of the hundreds column is Q or 13 (since Q = 0); the units digit of this sum does not match the 9 in the total. Thus, we conclude that Q cannot equal 0 and thus must equal 5. Verifying that Q = 5, we check the sum of the tens column again. Since = 19, then 1 is carried to the hundreds column. The sum of the hundreds column is = 19, as required. Thus, P + Q = = 14 and the completed addition is shown below We use labels, m and n, in the fourth row of the grid, as shown. Then, 10, m, 36, n are four terms of an arithmetic sequence. Since 10 and 36 are two terms apart in this sequence, and their difference is = 6, the constant added to one term to obtain the next term in the fourth row is 6 or 13. That is, m = = 3, and n = = 49. (We confirm that the terms 10, 3, 36, 49 do form an arithmetic sequence.) In the fourth column, 5 and n (which equals 49) are two terms apart in this sequence, and their difference is 49 5 = 4. Thus, the constant added to one term to obtain the next term in the fourth column is 4 or 1. That is, x = = 37 (or x = 49 1 = 37). The completed grid is as shown m Answer: (C) 5 x n Answer: (A) 3. Since P QR is isosceles with P Q = QR and P QR = 90, then QP R = QRS = 45. Also in P QR, altitude QS bisects P R (P S = SR) forming two identical triangles, SQP and SQR.

43 013 Gauss Contest s Page 8 Since these two triangles are identical, each has 1 of the area of P QR. In SQR, QSR = 90, QRS = 45, and so SQR = 45. Thus, SQR is also isosceles with SQ = SR. Then similarly, altitude ST bisects QR (QT = T R) forming two identical triangles, SQT and SRT. Since these two triangles are identical, each has 1 of the area of SQR or 1 of the area of 4 P QR. Continuing in this way, altitude T U divides ST R into two identical triangles, ST U and RT U. Each of these two triangles has 1 of 1 or 1 of the area of P QR. 4 8 Continuing, altitude UV divides RT U into two identical triangles, RUV and T UV. Each of these two triangles has 1 of 1 or 1 of the area of P QR Finally, altitude V W divides RUV into two identical triangles, UV W and RV W. Each of these two triangles has 1 of 1 or 1 of the area of P QR Since the area of ST U is 1 1 of the area of P QR, and the area of UV W is of the area 8 3 of P QR, then the total fraction of P QR that is shaded is = 4+1 or Answer: (D) 4. We begin by numbering the checkerboard squares from 1 to 16, as shown, so that we may refer to each of them specifically We denote a move up by the letter U and a move right by R We will begin by determining which of the 16 squares will not be touched by the face with the circle and then proceed to show that 8 3 the remaining squares will be touched by the face with the circle. up 4 Since the cube begins on square 1 and the circle is facing out, right square 1 will not be touched by the face with the circle. Similarly, each of the squares, 3, 4 can only be reached by moving the cube right (the sequence of moves to reach each of these three squares is R, RR and RRR, respectively), and in each case the circle remains facing out. Squares, 3 and 4 will not be touched by the face with the circle. Squares 5 and 9 can only be reached by moving the cube up (the sequence of moves to reach each of these two squares is U and UU, respectively). In either case, the face with the circle will not touch squares 5 and 9. Square 6 can be reached with two different sequences of moves, RU or UR. In both cases, the face with the circle will not touch square 6. Square 10 can be reached with three different sequences of moves, UUR, URU or RUU. In all three cases, the face with the circle will not touch square 10. In turns out that these eight squares (1,, 3, 4, 5, 6, 9, 10) are the only squares that will not be touched by the face with the circle on any path. The table below lists sequences of moves that demonstrate how each of the remaining eight squares will be touched by the face with the circle. The second column lists the sequence of moves, while the third column lists the position of the face with the circle as the cube progresses through the sequence of moves. We have used the letters F for front, B for back, T for top, O for bottom, L for left, and R for right to indicate the location of the face containing the circle.

44 013 Gauss Contest s Page 9 Square Sequence of Moves Position of the Circle 7 URR T RO 8 RURR F T RO 11 URUR T RRO 1 RURUR F T RRO 13 UUU T BO 14 RUUU F T BO 15 RRUUU F F T BO 16 RRRUUU F F F T BO Therefore, the number of different squares that will not be contacted by the face with the circle on any path is 8. Answer: (C) 5. We denote the number of tickets of each of the five colours by the first letter of the colour. We are given that b : g : r = 1 : : 4 and that g : y : o = 1 : 3 : 6. Through multiplication by, the ratio 1 : 3 : 6 is equivalent to the ratio : 6 : 1. Thus, g : y : o = : 6 : 1. We chose to scale this ratio by a factor of so that the only colour common to the two given ratios, green, now has the same number in both of these ratios. That is, b : g : r = 1 : : 4 and g : y : o = : 6 : 1 and since the term g is in each ratio, then we can combine these to form a single ratio, b : g : r : y : o = 1 : : 4 : 6 : 1. This ratios tells us that for every blue ticket, there are green, 4 red, 6 yellow, and 1 orange tickets. Thus, if there was only 1 blue ticket, then there would be = 5 tickets in total. However, we are given that the box contains 400 tickets in total. Therefore, the number of blue tickets in the box is 400 = Through multiplication by 16, the ratio b : g : r : y : o = 1 : : 4 : 6 : 1 becomes b : g : r : y : o = 16 : 3 : 64 : 96 : 19. (Note that there are = 400 tickets in total.) Next, we must determine the smallest number of tickets that must be drawn to ensure that at least 50 tickets of one colour have been selected. It is important to consider that up to 49 tickets of any one colour could be selected without being able to ensure that 50 tickets of one colour have been selected. That is, it is possible that the first 195 tickets selected could include exactly 49 orange, 49 yellow, 49 red, all 3 green, and all 16 blue tickets ( = 195). Since all green and blue tickets would have been drawn from the box, the next ticket selected would have to be the 50 th orange, yellow or red ticket. Thus, the smallest number of tickets that must be drawn to ensure that at least 50 tickets of one colour have been selected is 196. Answer: (D)

45 013 Gauss Contest s Page 10 Grade 8 1. Evaluating, = = = Evaluating, = 1 15 = 3. Answer: (D) Answer: (D) 3. 1 To determine the smallest number in the set { 1 common denominator of 1. The set { 1,, 1, 5, or to the set { 6, 8, 3, 10, }. The smallest number in this set is 3 1, so 1 4,, 1, 5, }, we express each number with a } { is equivalent to the set 1 6, 4, 1 3, 5, is the smallest number in the original set. With the exception of 1, each number in the set is greater than or equal to 1. 4 We can see this by recognizing that the numerator of each fraction is greater than or equal to one half of its denominator. Thus, 1 is the only number in the list that is less than 1 and so it must be the smallest number 4 in the set. Answer: (C) 4. Since Ahmed stops to talk with Kee one quarter of the way to the store, then the remaining distance to the store is 1 1 = 3 of the total distance. 4 4 Since 3 = 3 1, then the distance Ahmed travelled from Kee to the store, is 3 times the 4 4 distance Ahmed travelled from the start to reach Kee. That is, 1 km is 3 times the distance between the start and Kee. So the distance between the start and Kee is 1 = 4 km. 3 Therefore, the total distance travelled by Ahmed altogether is or 16 km. Answer: (B) 5. Since Jarek multiplies a number by 3 and gets an answer of 90, then the number must be 90 = 30. (We may check that 30 3 = 90.) 3 If Jarek instead divides the number 30 by 3, then the answer he gets is 30 = Answer: (B) } 6. 1 We first evaluate the product on the left side of the equation = = = = Similarly, we evaluate the product on the right side of the equation = = = The left side of the equation is equal to the right side of the equation, so then = Therefore, the number that goes in the box is = 50. The product of the first two numbers on the left side of the equation is equal to the product of the first two numbers on the right side of the equation. That is, 10 0 = 00 = 100. Similarly, the product of the next two numbers (the third and fourth numbers) on the left

46 013 Gauss Contest s Page 11 side of the equation is equal to the product of the next two numbers on the right side of the equation. That is, = 100 = Since the products of the first four numbers on each side of the equation are equal, then the final (fifth) number on each side of the equation must be equal. Therefore, the number that goes in the box is equal to the fifth number on the left side of the equation or 50. Answer: (C) 7. There are 6 letters in the English alphabet. There are 5 different letters, a, l, o, n, s, in Alonso s name. If one letter is randomly drawn from the bag, then the probability that it is a letter in Alonso s name is 5 6. Answer: (C) 8. When Mathy Manuel s autograph dropped 30% in value, it lost $ = $30 of its value. After this drop, the autograph was worth $100 $30 = $70. If the autograph then increased by 40% in value, the increase would be $ = $8. After this increase, the autograph would be worth $70 + $8 = $98. Answer: (A) 9. After reflecting the point (, 3) in the x-axis, the x-coordinate of the image will be the same as the x-coordinate of the original point, x =. The original point is a distance of 3 below the x-axis. The image will be the same distance from the x-axis, but above the x-axis. Thus, the image has y-coordinate 3. The coordinates of the image point are (, 3). (, 3) 4 (, 3) y x Answer: (E) 10. The value of four nickels is 4 5 = 0. The value of six dimes is 6 10 = 60. The value of two quarters is 5 = 50. The ratio of the value of four nickels to six dimes to two quarters is 0 : 60 : 50 = : 6 : 5. Answer: (B) 11. Substituting x = 4 into 3x + y we get 3(4) + y = 1 + y. Since this expression 1 + y is equal to 30, then y must equal 30 1 or 18. If y = 18, then y is 18 or 9. Answer: (E)

47 013 Gauss Contest s Page Using order of operations to evaluate, ( 3 ) 4 3 = = = 0. We first express 4 as and then use exponent rules to evaluate. ( 3 ) 4 3 = ( 3 ) ( ) 3 = 3 3 = 6 6 = 0 Answer: (A) 13. Two consecutive Summer Olympics are held 4 years apart. Each successive Summer Olympics requires an additional 4 years before it is held. We use this to summarize the minimum time required for the largest number of Summer Olympics to be held, as shown. Number of Summer Olympics Minimum Number of Years Apart From the table above, we see that at least 0 years are required to host 6 Summer Olympics, while only 16 years are required to host 5. (For example, the 5 olympics could be held in years 1, 5, 9, 13, and 17.) Therefore, the maximum number of Summer Olympics that can be held during an 18 year period is 5. Answer: (C) 14. Let s be the side length of the cube. The surface area of a cube is made up of 6 identical squares. Since the surface area is 54 cm, then each of the 6 squares has area (54 6) cm = 9 cm. The area of a square with side length s is s, so s = 9 or s = 9 = 3 cm. The volume of a cube is given by the product of its length, width and height, which are all equal to s, 3 cm. Thus the volume of the cube with surface area 54 cm is = 7 cm 3. Answer: (D) When is divided by 13 with the help of a calculator, we get = Since = 9997 and = 3, then we have a quotient of 769 and a remainder of 3. That is, = This is called a division statement. Similarly, we may determine the remainder when each of the five possible answers (the dividend) is divided by 13 (the divisor). We summarize this work in the table below.

48 013 Gauss Contest s Page 13 Answer Division Statement Remainder (A) = (B) = (C) = (D) = (E) = Of the five possible answers, only gives a remainder of 3 when divided by 13. When is divided by 13, the remainder is 3. Thus, adding any multiple of 13 to will give the same remainder, 3, upon division by 13. Of the five choices given, the only answer that differs from by a multiple of 13 is When is divided by 13, the remainder is also 3. Answer: (C) 16. Since it is equally likely that a child is a boy as it is that a child is a girl, then the probability that any child is a girl is 1. The probability of any child being born a girl is independent of the number or gender of any children already born into the family. That is, the probability of the second child being a girl is also 1, as is the probability of the third child being a girl. Therefore, the probability that all 3 children are girls is = 1. 8 Answer: (E) 17. Since P QRS is a rectangle, then P QR = 90. Since P QR = P QS + RQS, then 90 = (5x) + (4x) or 90 = 9x and so x = 10. In SRQ, SRQ = 90 and RQS = (4x) = 40. Therefore, QSR = y = = 50. So y = 50. Answer: (D) 18. Sally s answer is = 3 3 = 6 6 = 1. Jane s answer is = = = 13 6 = 1 6. The difference between Sally s answer and Jane s answer is or Answer: (B) 19. Since we seek the minimum number of colours that Serena can use to colour the hexagons, we first determine if it is possible for her to use only two colours (using only one colour is not possible). We will use the numbers 1,, 3 to represent distinct (different) colours. We begin by choosing any group of three hexagons in which each pair of hexagons share a side, as shown. We colour two of the hexagons shown with colours 1 and 1 (since they share a side). Each of these two coloured hexagons share a side with the third hexagon which therefore can not be coloured 1 or. Thus, the minimum number of colours that Serena can use is at least three.

49 013 Gauss Contest s Page 14 Next, we determine if the entire tiling can be coloured using only three colours. One possible colouring of the tiles that uses only three colours is shown to the right While other colourings of the tiles are possible, Serena is able to use only three colours and ensure that no two hexagons that 1 1 share a side are the same colour There are many nice patterns of the colours in this tiling Can you find a different colouring of the tiles that uses only three colours? Answer: (E) 0. 1 Suppose that the cost of one book, in dollars, is C. Then Christina has 3 4 C and Frieda has 1 C. Combining their money, together Christina and Frieda have 3 4 C + 1 C = 3 4 C + 4 C = 5 4 C. If the book was $3 cheaper, then the cost to buy one book would be C 3. If the cost of one book was C 3, then the cost to buy two at this price would be (C 3) or C 6. Combined, Christina and Frieda would have enough money to buy exactly two books at this reduced price. Thus, C 6 = 5 4 C. Solving, C 6 = 5 4 C C 5 4 C = = = 6 Since 3 of 8 is 6, then C = 8. 4 Therefore, the original price of the book is $8. We proceed by systematically trying the five multiple choice answers given. Initial cost Combined money for Reduced cost Number of books of the book Christina and Frieda of the book they may buy $4 3 4 of $4 + 1 $ of $ $1 3 4 of $1 + 1 $ of $ $8 3 4 of $8 + 1 of $4 = $3 + $ = $5 $1 5 1 = 5 of $16 = $1 + $8 = $0 $ = of $1 = $9 + $6 = $15 $ = of $10 = $ $5 = $1.50 $ = of $8 = $6 + $4 = $10 $ = We see that if the original price of the book is $8, Christina and Frieda are able to buy exactly two copies of the book at the reduced price. Thus, the initial cost of the book is $8. Answer: (E)

50 013 Gauss Contest s Page We use labels m and n in the first column, as shown in the top grid. Then, 5, m, n, 3 are four terms of an arithmetic sequence. Since 5 and 3 are three terms apart in this sequence, and their difference is 3 5 = 18, the constant added to one term to obtain the next term in the first column is 18 or 6. 3 That is, m = = 11, and n = = 17. (We confirm that the terms 5, 11, 17, 3 do form an arithmetic sequence.) 5 m n 3 x We use labels p and q in the second row, as shown in the middle grid. Then, 11, p, q, 111 are four terms of an arithmetic sequence. Since 11 and 111 are three terms apart in this sequence, and their difference is = 100, the constant added to one term to obtain the next term in the second row is 100 or That is, p = = 411, and q = = 811. (We confirm that the terms 11, 411, 811, 111 do form an arithmetic sequence.) p x q We use label r in the third row, as shown in the third grid. Then, 17, r, 1013 are three terms of an arithmetic sequence. Since 17 and 1013 are two terms apart in this sequence, and their difference is = 996, the constant added to one term to obtain the next term in the third row is 996 or 498. That is, r = = 515. (We confirm that the terms 17, 515, 1013 do form an arithmetic sequence.) Finally, in the second column the terms 411, 515, x are three terms of an arithmetic sequence. Since 411 and 515 are one term apart in this sequence, the constant added to one term to obtain the next term in the second column is = 104. Then, x = = 619. The completed grid is as shown r 1013 x Answer: (B). In F GH, F G = GH = x since they are both radii of the same circle. By the Pythagorean Theorem, F H = F G +GH = x +x, or F H = x, and so ( 8) = x or x = 8 and x = 4, so then x = (since x > 0). F G, GH and arc F H form a sector of a circle with centre G and radius GH. Since F GH = 90, which is 1 of 4 360, then the area of this sector is one quarter of the area of the circle with centre G and radius GH = F G =. The shaded area is equal to the area of sector F GH minus the area of F GH. The area of sector F GH is 1 4 π() or 1 π(4) or π. 4 The area of F GH is F G GH or, so. Therefore, the area of the shaded region is π. Answer: (A)

51 013 Gauss Contest s Page In the first race, when Azarah crossed the finish line, Charlize was 0 m behind or Charlize had run 80 m. Since Azarah and Charlize each travelled these respective distances in the same amount of time, then the ratio of their speeds is equal to the ratio of their distances travelled, or 100 : 80. Similarly in the second race, when Charlize crossed the finish line, Greg was 10 m behind or Greg had run 90 m. Since Charlize and Greg each travelled these respective distances in the same amount of time, then the ratio of their speeds is equal to the ratio of their distances travelled, or 100 : 90. Let A, C and G represent Azarah s, Charlize s and Greg s speeds, respectively. Then, A : C = 100 : 80 = 5 : 0 and C : G = 100 : 90 = 0 : 18. Therefore, A : C : G = 5 : 0 : 18 and A : G = 5 : 18 = 100 : 7. Over equal times, the ratio of their speeds is equal to the ratio of their distances travelled. Therefore, when Azarah travels 100 m, Greg travels 7 m. When Azarah crossed the finish line, Greg was = 8 m behind. In the first race, when Azarah crossed the finish line, Charlize was 0 m behind or Charlize had run 80 m. Since Azarah and Charlize each travelled these respective distances in the same amount of time, then the ratio of their speeds is equal to the ratio of their distances travelled, or 100 : 80. That is, Charlize s speed is 80% of Azarah s speed. Similarly, Greg s speed is 90% of Charlize s speed. Therefore, Greg s speed is 90% of Charlize s speed which is 80% of Azarah s speed, or Greg s speed is 90% of 80% of Azarah s speed. Since 90% of 80% is equivalent to = 0.7 or 7%, then Greg s speed is 7% of Azarah s speed. When Azarah ran 100 m (crossed the finish line), Greg ran 7% of 100 m or 7 m in the same amount of time. When Azarah crossed the finish line, Greg was = 8 m behind. Answer: (C) 4. The length of the longest side, z, is less than half of the perimeter 57. Thus, z < 57 or z < 8 1. Since z is an integer then z 8. When z = 8, x + y = 57 8 = 9. We list all possible values for x and y in the table below given that z = 8 and x < y < z. y x Systematically, we continue decreasing the value of z and listing all possible values for x and y. When z = 7, x + y = 57 7 = 30. y x When z = 6, x + y = 57 6 = 31. y x When z = 5, x + y = 57 5 = 3. y x

52 013 Gauss Contest s Page 17 When z = 4, x + y = 57 4 = 33. y x When z = 3, x + y = 57 3 = 34. y x When z =, x + y = 57 = 35. y x When z = 1, x + y = 57 1 = 36. y 0 19 x When z = 0, x + y = 57 0 = 37. y 19 x 18 The next smallest value for z is 19 and in this case x + y = = 38. However, if x + y = 38 then at least one of x or y must be 19 or larger. This is not possible since z = 19 and x < y < z. Therefore, 0 is the smallest possible value for z and we have listed the side lengths of all possible triangles above. Counting, we see that there are = 61 possible triangles that satisfy the given conditions. Answer: (B) 5. 1 Let b represent the number of boys initially registered in the class. Let g represent the number of girls initially registered in the class. When 11 boys transferred into the class, the number of boys in the class was b When 13 girls transferred out of the class, the number of girls in the class was g 13. The ratio of boys to girls in the class at this point was 1 : 1. That is, the number of boys in the class was equal to the number of girls in the class, or b + 11 = g 13 and so g = b + 4. Since there were at least 66 students initially registered in the class, then b + g 66. Substituting g = b + 4, b + g becomes b + (b + 4) = b + 4, and so b or b 4, so b 1. At this point, we may use the conditions that g = b + 4 and b 1 to determine which of the 5 answers given is not possible. Each of the 5 answers represents a possible ratio of b : g or b : (b + 4) (since g = b + 4). We check whether b : (b + 4) can equal each of the given ratios while satisfying the condition that b 1. b In (A), we have b : (b + 4) = 4 : 7 or or 7b = 4b + 96 and then 3b = 96, so b = 3. In (B), we have b : (b + 4) = 1 : or In (C), we have b : (b + 4) = 9 : 13 or b = 54. In (D), we have b : (b + 4) = 5 : 11 or b = 0. = 4 b+4 7 b = 1 b+4 b = 9 b+4 13 b = 5 b+4 11 or b = b + 4, so b = 4. or 13b = 9b + 16 and then 4b = 16, so or 11b = 5b + 10 and then 6b = 10, so

53 013 Gauss Contest s Page 18 In (E), we have b : (b + 4) = 3 : 5 or b b+4 = 3 5 or 5b = 3b + 7 and then b = 7, so b = 36. Thus, the only ratio that does not satisfy the condition that b 1 is 5 : 11 (since b = 0). As in 1, we establish the conditions that g = b + 4 and b 1. Again, the ratio b : g or b b then becomes (since g = b + 4). g b+4 Since b 1, then b can equal 1,, 3, 4,..., but the smallest possible value for b is 1. When b = 1, When b =, b becomes 1 = 1 b b becomes = b+4 +4 b becomes 3 = 3 b = When b = 3, b When b = 4, becomes 4 = 4 = 0.5. b As b continues to increase, the value of the ratio b b+4 or b g Can you verify this for yourself? Thus, the smallest possible value of the ratio b 1 is or g 45 Comparing this ratio to the 5 answers given, we determine that 5 continues to increase. 11 = 0.45 < 0.46 = 1 45, which is not possible. (You may also confirm that each of the other 4 answers given are all greater than 1 and are 45 obtainable, as in 1.) Therefore, the only ratio of boys to girls which is not possible is (D). Answer: (D)

54 The CENTRE for EDUCATION in MATHEMATICS and COMPUTING 01 Gauss Contests (Grades 7 and 8) Wednesday, May 16, 01 (in North America and South America) Thursday, May 17, 01 (outside of North America and South America) s 011 Centre for Education in Mathematics and Computing

55 01 Gauss Contest s Page 3 Grade 7 1. Evaluating, = = Written numerically, the number 33 million is Answer: (B) Answer: (D) 3. Each of the numbers 1,, 3, 4, 5, and 6 is equally likely to appear when the die is rolled. Since there are six numbers, then each has a one in six chance of being rolled. The probability of rolling a 5 is 1 6. Answer: (B) 4. A positive fraction increases in value as its numerator increases and also increases in value as its denominator decreases. Since the numerators of all five fractions are equal, then the largest of these is the fraction with the smallest denominator. The largest fraction in the set { 1, 1, 1, 1, } is 1. Answer: (A) 5. 1 The angle marked is vertically opposite the angle measuring 60. Since vertically opposite angles are equal, then the measure of the angle marked is also 60. The measure of a straight angle is 180. Together, the angle measuring 10 and the angle marked make up a straight angle. That is, 10 + = 180. Therefore, the angle marked is 60. Answer: (A) 6. Since 15 times the number is 300, then the number equals 300 divided by 15, or 0. Answer: (A) 7. Consider placing each of the two numbers from each answer on a number line. Numbers decrease in value as we move to the left along a number line. Since the first number listed in the pair must be less than the second, we want the first number to be to the left of the second number on the number line. The five possible answers are each given on the number lines shown. The last number line is the only one for which the first number listed ( 8) is positioned to the left of (is less than) the second number ( ) Answer: (E)

56 01 Gauss Contest s Page 4 8. Since Bailey scores on 6 of her 8 shots, then she misses on 8 6 = shots. If she misses on = 1 of her shots, then the percentage of shots that she does not score is % = 5%. 4 Answer: (E) 9. The number of visits to Ben s website from Monday to Friday can be read from the graph. These are: 300, 400, 300, 00, 00. The mean number of visits per day is found by adding these five totals and then dividing the sum by 5. Thus, the mean is ( ) 5 = or 80. The mean number of visits per day to Ben s website over the 5 days is between 00 and 300. Answer: (C) 10. The graph shows that the vehicle travels at a constant speed of 0 m/s. Travelling at 0 m/s, it will take the vehicle = 5 seconds to travel 100 metres. Answer: (E) 11. Since the four sides of a square are equal in length and the perimeter is 36 cm, then each side has length 36 = 9 cm. 4 The area of the square is the product of the length and width, which are each equal to 9 cm. Therefore, the area of the square, in cm, is 9 9 = 81. Answer: (B) 1. Since = 15 4 and = = 15 4, then answers (B) and (E) both simplify to Written as a mixed fraction, 15 4 is equal to Since 3.75 = 3 3 = 3 + 3, then answers (A) and (C) both simplify to 3 3 and thus are equivalent to Simplifying answer (D), 5 3 = 5 3 = Thus, 5 3 is not equal to Answer: (D) 13. Since P Q passes through centre O, then it is a diameter of the circle. Since QOT = 90, then P OT = = 90. Thus, the area of sector P OT is 90 = 1 or 5% of the area of the circle. Since the areas labelled R and S are equal, then each is 5% = 1.5% of the area of the circle. Therefore, a spin will stop on the shaded region 1.5% of the time. O P Q R S T Answer: (E) 14. To make the difference as large as possible, we make one number as large as possible and the other number as small as possible. The tens digit of a number contributes more to its value than its units digit. Thus, we construct the largest possible number by choosing 8 (the largest digit) to be its tens digit, and by choosing 6 (the second largest digit) to be the ones digit. Similarly, we construct the smallest possible number by choosing (the smallest digit) to be its tens digit, and 4 (the second smallest digit) to be its ones digit. The largest possible difference is 86 4 = 6. Answer: (B)

57 01 Gauss Contest s Page Since 1 mm of snow falls every 6 minutes, then 10 mm will fall every 6 10 = 60 minutes. Since 10 mm is 1 cm and 60 minutes is 1 hour, then 1 cm of snow will fall every 1 hour. Since 1 cm of snow falls every 1 hour, then 100 cm will fall every = 100 hours. Answer: (E) 16. Both 1 and 01 are obvious positive integer factors of 01. Since 01 is an even number and 01 = 1006, then both and 1006 are factors of 01. Since 1006 is also even then 01 is divisible by 4. Since 01 4 = 503, then both 4 and 503 are factors of 01. We are given that 503 is a prime number; thus there are no additional factors of 503 and hence there are no additional factors of 01. The factors of 01 are 1 and 01, and 1006, 4 and 503. Therefore, there are 6 positive integers that are factors of 01. Answer: (D) Since the ratio of boys to girls is 8 : 5, then for every 5 girls there are 8 boys. That is, the number of girls at Gauss Public School is 5 of the number of boys. 8 Since the number of boys at the school is 18, the number of girls is = = The number of students at the school is the number of boys added to the number of girls or = 08. Since the ratio of boys to girls is 8 : 5, then for every 8 boys there are = 13 students. That is, the number of students at Gauss Public School is 13 of the number of boys. 8 Since the number of boys at the school is 18, the number of students is = = Answer: (C) 18. In turn, we may use each of the three known scales to find a way to balance a circle, a diamond and a triangle. Since many answers are possible, we must then check our solution to see if it exists among the five answers given. First consider the scale at the top right. A diamond and a circle are balanced by a triangle. If we were to add a triangle to both sides of this scale, then it would remain balanced and the right side of this scale would contain what we are trying to balance, a circle, a diamond and a triangle. That is, a circle, a diamond and a triangle are balanced by two triangles. However, two triangles is not one of the five answers given. Next, consider the scale at the top left. A triangle and a circle are balanced by a square. If we were to add a diamond to both sides of this scale, then it would remain balanced and the left side of this scale would contain what we are trying to balance, a circle, a diamond and a triangle. That is, a circle, a diamond and a triangle are balanced by a square and a diamond. This answer is given as one of the five answers. In the context of a multiple choice contest, we do not expect that students will verify that the other four answers do not balance a circle, a diamond and a triangle. However, it is worth noting that it can be shown that they do not. Answer: (D)

58 01 Gauss Contest s Page In an ordered list of five integers, the median is the number in the middle or third position. Thus if we let the set of integers be a, b, c, d, e, ordered from smallest to largest, then c = 18. Since the average is fixed (at 0), e (the largest number in the set) is largest when a, b and d are as small as possible. Since the numbers in the set are different positive integers, the smallest that a can be is 1 and the smallest that b can be is. Our set of integers is now 1,, 18, d, e. Again, we want d to be as small as possible, but it must be larger than the median 18. Therefore, d = 19. Since the average of the 5 integers is 0, then the sum of the five integers is 5 0 = 100. Thus, e = 100 or 40 + e = 100, and so e = 60. The greatest possible integer in the set is 60. Answer: (A) 0. If either Chris or Mark says, Tomorrow, I will lie. on a day that he tells a lie, then it actually means that tomorrow he will tell the truth (since he is lying). This can only occur when he lies and then tells the truth on consecutive days. For Chris, this only happens on Sunday, since he lies on Sunday but tells the truth on Monday. For Mark, this only happens on Thursday, since he lies on Thursday but tells the truth on Friday. Similarly, if either Chris or Mark says, Tomorrow, I will lie. on a day that they tell the truth, then it means that tomorrow they will lie (since they are telling the truth). This can only occur when they tell the truth and then lie on consecutive days. For Chris, this only happens on Thursday, since he tells the truth on Thursday but lies on Friday. For Mark, this only happens on Monday, since he tells the truth on Monday but lies on Tuesday. Therefore, the only day of the week that they would both say, Tomorrow, I will lie., is Thursday. Answer: (B) 1. The triangular prism given can be created by slicing the 3 cm by 4 cm base of a rectangular prism with equal height across its diagonal. That is, the volume of the triangular prism in question is one half of the volume of the rectangular prism shown. Since the volume of the triangular prism is 10 cm 3, then the volume of this rectangular prism is 10 = 40 cm 3. The volume of the rectangular prism equals the area of its base 3 4 times its height, h. Since the volume is 40, then 3 4 h = 40 or 1h = 40, so h = 40 = 0. 1 Since the height of this rectangular prism is equal to the height of the triangular prism in question, then the required height is 0 cm h Answer: (B)

59 01 Gauss Contest s Page 7. Without changing the overall class mean, we may consider that the class has 100 students. That is, 0 students got 0 questions correct, 5 students got 1 question correct, 40 students got questions correct, and 35 students got 3 questions correct. The combined number of marks achieved by all 100 students in the class is then, (0 0) + (5 1) + (40 ) + (35 3) = = 190. Since the 100 students earned a total of 190 marks, then the overall class average was = 1.9. Answer: (B) 3. The units digit of any product is given by the units digit of the product of the units digits of the numbers being multiplied. For example, the units digit of the product 1 53 is given by the product 3, so it is 6. Thus to determine the units digit of N, we need only consider the product of the units digits of the numbers being multiplied to give N. The units digits of the numbers in the product N are 1, 3, 7, 9, 1, 3, 7, 9,..., and so on. That is, the units digits 1, 3, 7, 9 are repeated in each group of four numbers in the product. There are ten groups of these four numbers, 1, 3, 7, 9, in the product. We first determine the units digit of the product The units digit of 1 3 is 3. The units digit of the product 3 7 is 1 (since 3 7 = 1). The units digit of 1 9 is 9. Therefore, the units digit of the product is 9. (We could have calculated the product = 189 to determine the units digit.) This digit 9 is the units digits of the product of each group of four successive numbers in N. Thus, to determine the units digit of N we must determine the units digit of This product is equal to Since we are multiplying numbers with units digit 1, then the units digit of the product is 1. Answer: (A) 4. Diagonal P R divides parallelogram P QRS into two equal areas. That is, the area of P RS is one half of the area of parallelogram P QRS, or 0. In P RS, we construct median RT. (A median is a line segment that joins a vertex of a triangle to the midpoint of its opposite side.) Median RT divides P RS into two equal areas since its base, P S, is halved while the height remains the same. That is, the area of T RS is one half of the area of P RS, or 10. P Q T S V R Similarly, we construct median T V in T RS, as shown. Median T V divides T RS into two equal areas since its base, SR, is halved while the height remains the same. That is, the area of T V S is one half of the area of T RS, or 5. The area of P RV T is equal to the area of T V S subtracted from the area of P RS, or 0 5 = 15. P Q T R V S Answer: (C)

60 01 Gauss Contest s Page 8 5. A very useful and well-known formula allows us to determine the sum of the first n positive integers, (n 1) + n. n(n + 1) The formula says that this sum, (n 1)+n, is equal to (justification of this formula is included at the end of the solution). 6(6 + 1) For example, if n = 6 then = = 6 7 = 4 = 1. You can check that this formula gives the correct sum, 1, by mentally adding the positive integers from 1 to 6. In the table given, there is 1 number in Row 1, there are numbers in Row, 3 numbers in Row 3, and so on, with n numbers in Row n. The numbers in the rows list the positive integers in order beginning at 1 in Row 1, with each new row containing one more integer than the previous row. Thus, the last number in each row is equal to the sum of the number of numbers in the table up to that row. For example, the last number in Row 4 is 10, which is equal to the sum of the number of numbers in rows 1,, 3, and 4. But the number of numbers in each row is equal to the row number. So 10 is equal to the sum That is, the last number in Row n is equal to the sum (n 1) + n, which n(n + 1) is equal to. We may now use this formula to determine in what row the number 000 appears. Using trial and error, we find that since 6(63) = 1953, then the last number in Row 6 is Similarly, since 63(64) = 016, then the last number in Row 63 is 016. Since 000 is between 1953 and 016, then 000 must appear somewhere in Row 63. To find how many integers less than 000 are in the column that contains the number 000, we must determine in which column the number 000 appears. Further, we must determine how many numbers there are in that column above the 000 (since all numbers in that column in rows below the 63 rd are larger than 000). We know that 016 is the last number in Row 63 and since it is the last number, it will have no numbers in the column above it. Moving backward (to the left) from 016, the number 015 will have 1 number in the column above it, 014 will have numbers in the column above it, and so on. That is, if we move k numbers to the left of 016, that table entry will have k numbers in the column above it. In other words, if the number 016 k appears in Row 63, then there are k integers less than it in the column that contains it. Since we know that 000 appears in this 63 rd row, then 016 k = 000 means that k = 16. Thus, there are 16 integers less than 000 in the column that contains the number 000. n(n + 1) Verification of the Formula: (n 1) + n = If we let the sum of the first n positive integers be S, then S = (n 1) + n. If this same sum is written in the reverse order, then S = n+(n 1)+(n )+(n 3)+ ++1.

61 01 Gauss Contest s Page 9 Adding the right sides of these two equations, (n 1) + n + n + (n 1) + (n ) + (n 3) = (n + 1) + (n + 1) + (n + 1) + (n + 1) (n + 1) + (n + 1) In this sum there are n occurrences of (n + 1), hence the sum is n(n + 1). n(n + 1) However, this sum represents S + S or S, so if S = n(n + 1) then S =. Answer: (D)

62 01 Gauss Contest s Page 10 Grade 8 1. Using the correct order of operations, 3 (3 + 3) 3 = = 18 3 = 6. Answer: (A). Each of the numbers 1,, 3, 4, 5, and 6 are equally likely to appear when the die is rolled. Since there are 6 numbers, then each has a one in six chance of being rolled. The probability of rolling a five is 1 6. Answer: (B) 3. Written first as a fraction, fifty-six hundredths is The decimal equivalent of this fraction can be determined by dividing, = = Answer: (D) 4. Since P, Q, R lie in a straight line, P QR = 180. Therefore, 4 + x + x = 180 or 4 + x = 180 or x = = 138, and so x = 69. Answer: (A) 5. The number of 5 coins needed to make one dollar (100 ) is = 0. The number of 10 coins needed to make one dollar (100 ) is = 10. Therefore, it takes 0 10 = 10 more 5 coins than it takes 10 coins to make one dollar. Answer: (B) 6. Once each of 1 equal parts is cut into equal pieces, there are 1 = 4 equal pieces of pizza. Ronald eats 3 of these 4 equal pieces. Therefore, Ronald eats 3 4 or 1 8 of the pizza. Answer: (E) 7. Since the rectangular sheet of paper measures 5 cm by 9 cm, its area is 5 9 or 5 cm. A square sheet of paper has equal length and width. If the length and width of the square is s, then the area of the square is s s or s. Therefore s = 5, or s = 5 = 15 (since s is a positive length). The dimensions of the square sheet of paper having the same area are 15 cm by 15 cm. Answer: (A) 8. Since the number in question (0.01) is in decimal form, it is easiest to determine into which of the 5 given ranges it falls by converting the ranges into decimal form also. 1 Converting, is 0.1, 1 is 0., 1 is 0.5, 1 is 0.3, and 1 is Since 0.01 is greater than 0. but less than 0.5, it is between 1 and Answer: (C) 9. Substituting x =, we get 3 x x 3 = 3 3 = (3 3) ( ) = 9 8 = 1. Answer: (D)

63 01 Gauss Contest s Page The area of the rectangle is 8 4, or 3. 8 The unshaded portion of the rectangle is a triangle with base of length 8 and height h, as shown. Since the dotted line (the height) with length h is parallel to the vertical side of the rectangle, then h = 4. h 4 Thus, the area of the unshaded triangle is = 4 4 = 16. The area of the shaded region is the area of the rectangle minus the area of the unshaded triangle. Thus, the area of the shaded region is 3 16 = 16. Answer: (B) 11. Since the pyramid has a square base, the base of the pyramid has 4 edges (one for each side of the square). An edge joins each of the 4 vertices of the square to the apex of the pyramid, as shown. In total, a pyramid with a square base has 8 edges. Answer: (A) 1. Since 1 mm of snow falls every 6 minutes, then 10 mm will fall every 6 10 = 60 minutes. Since 10 mm is 1 cm and 60 minutes is 1 hour, then 1 cm of snow will fall every 1 hour. Since 1 cm of snow falls every 1 hour, then 100 cm will fall every = 100 hours. Answer: (E) 13. The mode is the number that occurs most frequently in a set of numbers. The mode of the three numbers is 9. Thus, at least two of the three numbers must equal 9 otherwise there would be three different numbers and so three modes. If all three of the numbers were equal to 9, then the average of the three numbers could not be 7. Therefore, two of the three numbers are equal to 9. If the third number is x, then the three numbers are x, 9 and 9. Since the average of the three numbers is 7, then x = 7 or x + 18 = 1, so x = 3. 3 The smallest of the three numbers is 3. Answer: (C) Since half the square root of the number is one, then the square root of the number must be. Since the square root of 4 is equal to, then the unknown number is 4. Let the unknown number be x. Since one half the square root of x is one, then 1 x = 1. Multiplying both sides of the equation by gives, 1 x = 1 or x =. Squaring both sides of the equation gives, ( x) = or x = 4. Answer: (B) 15. To determine which combination will not be said, we list the letters and numbers recited by Yelena and Zeno in the table below.

64 01 Gauss Contest s Page 1 Yelena P Q R S T U P Q R S T U Zeno Recognize that at this point Yelena has just said U and thus will begin from the beginning, P, again. Similarly, Zeno has just said 4 and thus will begin from the beginning, 1, again. That is, the sequence of letter and number combinations listed above will continue to repeat. To determine which combination will not be said we need only compare the 5 answers with the 1 possibilities given in the table. The only combination that does not appear in the table and thus that will not be said is R. Answer: (D) Since the lot has 5% more cars than trucks, and since we are asked to determine the ratio of cars to trucks, we may begin by choosing a convenient number of trucks in the lot. Assume that there are 100 trucks in the lot. Since there are 5% more cars than trucks, and since 5% of 100 is 5, then there are 15 cars in the parking lot. Therefore, the ratio of cars to trucks in the parking lot is 15 : 100 or 5 : 4. Assume that the number of trucks in the parking lot is x. Since there are 5% more cars than trucks in the lot, then the number of cars is 1.5x. The ratio of cars to trucks is 1.5x : x or 1.5 : 1 or (1.5 4) : (1 4) = 5 : 4. Answer: (D) 17. The tens digit of a number contributes more to its value than its units digit. Thus in order to make the difference between the numbers as small as possible, we begin by making the difference between the two numbers tens digits as small as possible. The smallest possible difference between the two tens digits is and this is achieved in three ways only. That is, we may allow the tens digits to be and 4, or 4 and 6, or 6 and 8, as shown below Next, we complete the numbers by using the two digits that remain as the ones digits. We continue to make the difference as small as possible by using the larger of the two remaining digits to complete the smaller number, and using the smaller of the two remaining digits to complete the larger number. The completion of the three possible cases as well as their respective differences are shown below The smallest possible difference is 6 48 = 14. Answer: (B)

65 01 Gauss Contest s Page The triangular prism given can be created by slicing the 3 cm by 4 cm base of a rectangular prism with equal height across its diagonal. That is, the volume of the triangular prism in question is one half of the volume of the rectangular prism shown. Since the volume of the triangular prism is 10 cm 3, then the volume of this rectangular prism is 10 = 40 cm 3. The volume of the rectangular prism equals the area of its base 3 4 times its height, h. Since the volume is 40, then 3 4 h = 40 or 1h = 40, so h = 40 = 0. 1 Since the height of this rectangular prism is equal to the height of the triangular prism in question, then the required height is 0 cm h Answer: (B) 19. Since there are 480 student participants and each student is participating in 4 events, then across all events the total number of (non-unique) participants is = 190. Each event has 0 students participating. Thus, the number of different events is = 96. Each event is supervised by 1 adult coach, and there are 16 adult coaches each supervising the same number of events. Therefore, the number of events supervised by each coach is = 6. Answer: (C) 0. 1 Since the probability that Luke randomly chooses a blue marble is, then for every blue 5 marbles in the bag there are 5 marbles in total. Since there are only red and blue marbles in the bag, then for every blue marbles, there are 5 or 3 red marbles. Thus, if there are k blue marbles in the bag then there are 3k red marbles in the bag (where k is some positive integer). If Luke adds 5 blue marbles to the bag, then the number of blue marbles in the bag is k + 5. If in addition to adding 5 blue marbles to the bag Luke removes 5 red marbles from the bag, then the total number of marbles in the bag, k + 3k, remains the same. The probability that Luke randomly chooses a blue marble from the bag is determined by dividing the number of blue marbles in the bag, k + 5, by the total number of marbles in the bag, 5k. Thus, k + 5 = 3 or 5(k + 5) = 3(5k). 5k 5 Dividing both sides of this equation by 5 we have, k + 5 = 3k and so k = 5. Therefore, the total number of marbles in the bag is 5k = 5(5) = 5. Let the initial number of marbles in the bag be N. Since the probability that Luke randomly chooses a blue marble from the bag is, then there 5 are N blue marbles initially. 5 When 5 blue marbles are added to the bag and 5 red marbles are removed from the bag, the number of marbles in the bag is still N. Since the probability of choosing a blue marble at the end is 3, then there are 3 N blue marbles 5 5 in the bag at the end.

66 01 Gauss Contest s Page 14 The difference between the number of blue marbles initially and the number of blue marbles at the end is 5, since 5 blue marbles were added. Thus, 3N N = 5 or 1 N = 5 or N = Therefore, the total number of marbles in the bag is 5. Answer: (E) 1. From the first scale, 1 circle balances triangles. If we double what is on both sides of this scale, then circles balance 4 triangles. From the second scale, circles also balance 1 triangle and 1 square. So 4 triangles must balance 1 triangle and 1 square, or 3 triangles must balance 1 square. If 3 triangles balance 1 square, then 6 triangles balance squares. (We note at this point that although we have found a way to balance squares, 6 triangles is not one of the five answers given and so we continue on.) Six triangles is equivalent to 4 triangles plus triangles and we know from our doubling of the first scale that 4 triangles is balanced by circles. So squares is balanced by 6 triangles, which is 4 triangles plus triangles, which is balanced by circles and triangles. Therefore, a possible replacement for the? is circles and triangles. Answer: (D). Since only one of the given statements is true, then the other two statements are both false. Assume that the second statement is true (then the other two are false). If the second statement is true, then The Euclid is the tallest building. Since the third statement is false, then The Galileo is the tallest building. However, The Euclid and The Galileo cannot both be the tallest building so we have a contradiction. Therefore the second statement cannot be true. Assume that the third statement is true. Then The Galileo is not the tallest building. Since the second statement must be false, then The Euclid is not the tallest building. If both The Galileo and The Euclid are not the tallest building, then The Newton must be the tallest. However, since the first statement must also be false, then The Newton is the shortest building and we have again reached a contradiction. Therefore, the third statement cannot be true. Since both the second and third statements cannot be true, then the first statement must be true. Since the first statement is true, then The Newton is either the second tallest building or the tallest building. Since the third statement is false, then The Galileo is the tallest building which means that The Newton is the second tallest. Since the second statement is false, then The Euclid is not the tallest and therefore must be the shortest (since The Newton is second tallest). Ordered from shortest to tallest, the buildings are The Euclid (E), The Newton (N), and The Galileo (G). Answer: (C)

67 01 Gauss Contest s Page Care is needed in systematically counting the different patterns. Dividing the patterns into groups having some like attribute is one way to help this process. We will count patterns by grouping them according to the number of corner triangles (3,, 1, or 0) that each has shaded. 3 shaded corners There is only one pattern having all 3 corners shaded, as shown in Figure 1. shaded corners To start, we fix the corners that are shaded (the top and bottom right) since rotations will give the other two possibilities that have shaded 4 3 corners. Figure A For the purpose of identifying the smaller triangles, they have been numbered 1 to 6 as shown in Figure A. We need only shade one of these 6 numbered Figure Figure 3 Figure 4 triangles to complete a pattern. Since triangles and 6 each share a side with a shaded corner, they cannot be shaded. Shading triangle 1 gives our first pattern in this group, as shown in Figure. Triangles 3 and 4 are then shaded to give Figures 3 and 4, respectively. Note that shading triangle 5 gives a reflection of Figure 3 and so is not a different pattern. These 3 patterns shown cannot be matched by rotations or reflections. Thus, there are 3 different patterns having shaded corners. Figure 1 1 shaded corner Again we start by fixing the corner that is shaded (the top) since rotations will give the other two possibilities that have 1 shaded corner. 3 We identify the smaller triangles by number in Figure B Figure B. We need to shade of these 6 numbered triangles to complete a pattern. Since triangle 6 shares a side with the shaded Figure 5 Figure 6 Figure 7 corner, it cannot be shaded. Also, no two adjacent triangles can be shaded since they share a side. Shading triangles 1 and 3 gives our first pattern in this group, as shown in Figure 5. Triangles 1 and 4 are then shaded to give Figure 6. Figure 7 has triangles 1 and 5 shaded, while Figure 8 has triangles and 4 shaded. Shading triangles and 5 gives a reflection of Figure 6 and so is not a different pattern. Shading triangles 3 and 5 gives a reflection of Figure 5 and so is not a different pattern. These 4 patterns shown cannot be matched by rotations or reflections. There are no other combinations of triangles that can be shaded. Thus, there are 4 different patterns having 1 shaded corner. Figure 8

68 01 Gauss Contest s Page 16 No shaded corners We identify the smaller triangles by number in Figure C. We need to shade 3 of these 6 numbered triangles to complete a pattern. Since no two adjacent triangles can be shaded there are only two possible patterns. Triangles 1, 3 and 5 can be shaded (Figure 9) or triangles, 4 and 6 can be shaded (Figure 10). These patterns shown cannot be matched by rotations or reflections. There are no other combinations of 3 triangles that can be shaded. Thus, there are different patterns having no shaded corners. Since each of the 4 groups of patterns above has a different number of corner triangles shaded, there is no pattern in any one group that can be matched to a pattern from another group by rotations or reflections. Therefore, the total number of patterns that can be created is = 10. Answer: (C) 4. We must find all possible groups of stones that can be selected so that the group sum is 11. To do this, we first consider groups of size two. There are 5 groups of size two whose numbers sum to 11. These groups are: {1, 10}, {, 9}, {3, 8}, {4, 7}, and {5, 6}. Next, we consider groups of size three. There are 5 groups of size three whose numbers sum to 11. These are: {1,, 8}, {1, 3, 7}, {1, 4, 6}, {, 3, 6}, and {, 4, 5}. Can you verify that there are no other groups of size three? Although a group of size four is possible, ({1,, 3, 5}), it is not possible to form two more groups whose numbers sum to 11 using the remaining stones ({4, 6, 7, 8, 9, 10}). Therefore, only groups of size two and size three need to be considered. Next, we must count the number of ways that we can select three of the ten groups listed above such that no number is repeated between any of the three groups. We may consider four cases or ways that these 3 groups can be chosen. They are: all 3 groups are of size two, groups are of size two and 1 group is of size three, 1 group is of size two and groups are of size three, and finally all 3 groups are of size three. Case 1: all 3 groups are of size two Since there is no repetition of numbers between any of the 5 groups having size two, choosing any 3 groups will produce a solution. There are 10 possible solutions using groups of size two only. These are: Figure 9 {1, 10}, {, 9}, {3, 8} {1, 10}, {4, 7}, {5, 6} {1, 10}, {, 9}, {4, 7} {, 9}, {3, 8}, {4, 7} {1, 10}, {, 9}, {5, 6} {, 9}, {3, 8}, {5, 6} {1, 10}, {3, 8}, {4, 7} {, 9}, {4, 7}, {5, 6} {1, 10}, {3, 8}, {5, 6} {3, 8}, {4, 7}, {5, 6} Case : groups are of size two and 1 group is of size three Since there is repetition of numbers between some of the groups of size two and some of the groups of size three, we need to take care when making our choices Figure C Figure 10

69 01 Gauss Contest s Page 17 To systematically work through all possible combinations, we will first choose a group of size three and then consider all pairings of groups of size two such that there is no repetition of numbers in either group of size two with the numbers in the group of size three. This work is summarized in the table below. 1 group of size three groups of size two {1,, 8} {4, 7}, {5, 6} {1, 3, 7} {, 9}, {5, 6} {1, 4, 6} {, 9}, {3, 8} {, 3, 6} {1, 10}, {4, 7} {, 4, 5} {1, 10}, {3, 8} There are 5 possible solutions using groups of size two and 1 group of size three. Case 3: 1 group is of size two and groups are of size three There is considerable repetition of numbers between the 5 groups of size three, {1,, 8}, {1, 3, 7}, {1, 4, 6}, {, 3, 6}, and {, 4, 5}. In fact, there are only two groups whose numbers have no repetition. These are {1, 3, 7} and {, 4, 5}. Can you verify that this is the only possibility? At this point, we are unable to choose a group of size two such there is no repetition with the groups {1, 3, 7} and {, 4, 5}. Therefore, there are no solutions using 1 group of size two and groups of size three. Case 4: all 3 groups are of size three In Case 3, we found that there were only groups of size three that had no repetition of numbers. Therefore it is not possible to find 3 groups of size three without repeating numbers. That is, Case 4 produces no solutions. Thus, the total number of possible arrangements of the stones into three groups, each having a sum of 11, is = 15. Answer: (E) 5. Since W XY Z is a rectangle, then XW Z = W ZY = 90. Also, W X = ZY = 15 and W Z = XY = 9. Thus, P W S and SZR are both right-angled triangles. In P W S, P S = 3 + 4, by the Pythagorean Theorem. Therefore, P S = = 5, and so P S = 5 = 5 (since P S > 0). Similarly in SZR, SR = 5 + 1, by the Pythagorean Theorem. Therefore, SR = = 169, and so SR = 169 = 13 (since SR > 0). In triangles P W S and RY Q, P W = RY = 3, W S = Y Q = 4, and P W S = RY Q = 90. Therefore, the area of P W S is equal to = 6, as is the area of RY Q equal to 6. In triangles SZR and QXP, SZ = QX = 5, ZR = XP = 1, and SZR = QXP = 90. Therefore, the area of SZR is equal to = 30, as is the area of QXP equal to 30. The area of parallelogram P QRS is determined by subtracting the areas of triangles P W S, RY Q, SZR, and QXP from the area of rectangle W XY Z. The area of rectangle W XY Z is W X XY or 15 9 = 135. Therefore, the area of parallelogram P QRS is 135 ( 6) ( 30) = 63. The area of parallelogram P QRS is determined by multiplying the length of its base by its perpendicular height.

70 01 Gauss Contest s Page 18 Let the base of P QRS be SR and thus the perpendicular height is P T. That is, SR P T = 63, or 13 P T = 63, so P T = Since ST P = P T R = 90, then P T S is a right-angled triangle. In P T S, ST = P S P T = 5 ( 63, 13) by the Pythagorean Theorem. Therefore, ST = = = 56, and so ST = 56 = 16 (since ST > 0) Answer: (D)

71 The CENTRE for EDUCATION in MATHEMATICS and COMPUTING 011 Gauss Contests (Grades 7 and 8) Wednesday, May 11, 011 s 010 Centre for Education in Mathematics and Computing

72 011 Gauss Contest s Page 3 Grade 7 1. Evaluating, = = = 8 1 = 7.. We must first add 9 and 16. Thus, = 5 = 5. Answer: (E) Answer: (E) 3. Reading from the bar graph, only 1 student chose spring. Since 10 students were surveyed, then the percentage of students that chose spring was 1 100% 10 or 10%. Answer: (B) 4. Since ground beef sells for $5.00 per kg, then the cost of 1 kg is $ = $ Answer: (C) 5. Since each of the numbers is between 1 and, we consider the tenths digits. The numbers , and are between 1 and 1.1, while and are both greater than 1.1. We may eliminate answers (D) and (E). Next, we consider the hundreths digits. While and each have a 1 as their hundreths digit, has a 0 and is therefore the smallest number in the list. The ordered list from smallest to largest is {1.0011, , , , } Answer: (B) 6. Since you randomly choose one of the five answers, then each has an equally likely chance of being selected. Thus, the probability that you select the one correct answer from the five is 1 5. Answer: (A) 7. Since we are adding 1 3 seven times, then the result is equal to Answer: (E) 8. Keegan paddled 1 km of his 36 km trip before lunch. Therefore, Keegan has 36 1 = 4 km left to be completed after lunch. The fraction of his trip remaining to be completed is 4 36 = 3. Answer: (D) 9. After reflecting the point (3, 4) in the x-axis, the x-coordinate of the image will be the same as the x-coordinate of the original point, x = 3. The original point is a distance of 4 from the x-axis. The image will be the same distance from the x-axis, but below the x-axis. Thus, the image has y-coordinate 4. The coordinates of the image point are (3, 4). y (3, 4) (3, - 4) x Answer: (D)

73 011 Gauss Contest s Page Anika said that the plant was a red rose. Cathy said that the plant was a red dahlia. If red was not the correct colour of the plant, then Anika was incorrect about the colour and therefore must be correct about the type; in other words, the plant is a rose. Similarly, if red was not the correct colour of the plant, then Cathy was incorrect about the colour and therefore must be correct about the type; in other words, the plant is a dahlia. But the plant cannot be both a rose and a dahlia. Therefore, Cathy and Anika must have been correct about the colour being red. Bill said that the plant was a purple daisy. Since we know that the colour of the plant is red, then Bill was incorrect about it being purple. Therefore, Bill must have been correct about it being a daisy. Thus, the plant is a red daisy. Answer: (E) 11. The angles x and 3x shown are complementary and thus add to 90. Therefore, x + 3x = 90 or 5x = 90 and so x = 90 5 = 18. Answer: (D) 1. Since the four sides of a square are equal in length and the perimeter is 8, then each side has length 8 4 = 7. The area of the square is the product of the length and width, which are each equal to 7. Therefore, the area of the square in cm is 7 7 = 49. Answer: (D) 13. Since Kayla ate less than Max and Chris ate more than Max, then Kayla ate less than Max who ate less than Chris. Brandon and Tanya both ate less than Kayla. Therefore, Max ate the second most. Answer: (D) 14. The smallest three digit palindrome is 101. The largest three digit palindrome is 999. The difference between the smallest three digit palindrome and the largest three digit palindrome is = 898. Answer: (B) 15. Since 10 minutes is equivalent to = 1 6 of an hour, the skier travels 1 6 = km. Answer: (C) 16. Any number of cm rods add to give a rod having an even length. Since we need an odd length, 51 cm, then we must combine an odd length from the 5 cm rods with the even length from the cm rods to achieve this. An odd length using 5 cm rods can only be obtained by taking an odd number of them. All possible combinations are shown in the table below. Number of 5 cm rods Length in 5 cm rods Length in cm rods Number of cm rods = = = = = 6 6 = = = = 6 6 = 3

74 011 Gauss Contest s Page 5 Note that attempting to use 11 (or more) 5 cm rods gives more than the 51 cm length required. Thus, there are exactly 5 possible combinations that add to 51 cm using 5 cm rods first followed by cm rods. Answer: (A) Choosing one meat and one fruit, the possible lunches are beef and apple, beef and pear, beef and banana, chicken and apple, chicken and pear, or chicken and banana. Of these 6 lunches, of them include a banana. Thus when randomly given a lunch, the probability that it will include a banana is 6 or 1 3. Each of the possible lunches that Braydon may receive contain exactly one fruit. The meat chosen for each lunch does not affect what fruit is chosen. Thus, the probability that the lunch includes a banana is independent of the meat that is served with it. Since there are 3 fruits to choose from, then the probability that the lunch includes a banana is 1 3. Answer: (A) In kilograms, let the weights of the 3 pumpkins in increasing order be A, B and C. The lightest combined weight, 1 kg, must come from weighing the two lightest pumpkins. That is, A + B = 1. The heaviest combined weight, 15 kg, must come from weighing the two heaviest pumpkins. That is, B + C = 15. Then the third given weight, 13 kg, is the combined weight of the lightest and heaviest pumpkins. That is, A + C = 13. Systematically, we may try each of the 5 possible answers. If the lightest pumpkin weighs 4 kg (answer (A)), then A + B = 1 gives 4 + B = 1, or B = 8. If B = 8, then 8 + C = 15 or C = 7. Since C represents the weight of the heaviest pumpkin, C cannot be less than B and therefore the lightest pumpkin cannot weigh 4 kg. If the lightest pumpkin weighs 5 kg (answer (B)), then A + B = 1 gives 5 + B = 1, or B = 7. If B = 7, then 7 + C = 15 or C = 8. If A = 5 and C = 8, then the third and final equation A + C = 13 is also true. Therefore, the weight of the smallest pumpkin must be 5 kg. In kilograms, let the weights of the 3 pumpkins in increasing order be A, B and C. The lightest combined weight, 1 kg, must come from weighing the two lightest pumpkins. That is, A + B = 1. The heaviest combined weight, 15 kg, must come from weighing the two heaviest pumpkins. That is, B + C = 15. Then the third given weight, 13 kg, is the combined weight of the lightest and heaviest pumpkins. That is, A + C = 13. Since A + B = 1 and A + C = 13, then C is one more than B. Since B + C = 15 and C is one more than B, then B = 7 and C = 8.

75 011 Gauss Contest s Page 6 Since A + B = 1, then A + 7 = 1 or A = 5. When A = 5, B = 7 and C = 8, the weights of the pairs of pumpkins are 1 kg, 13 kg and 15 kg as was given. Therefore, the weight of the lightest pumpkin is 5 kg. Answer: (B) 19. If each of the four numbers is increased by 1, then the increase in their sum is 4. That is, these four new numbers when added together have a sum that is 4 more than their previous sum T, or T + 4. This new sum T + 4 is now tripled. The result is 3 (T + 4) = (T + 4) + (T + 4) + (T + 4) or 3T + 1. Answer: (C) 0. The volume of the rectangular prism equals the area of its base 6 4 times its height. That is, the rectangular prism has a volume of 6 4 = 48 cm 3. The volume of the triangular prism is found by multiplying the area of one of its triangular faces by its length. The triangular face has a base of length 6 3 = 3 cm. This same triangular face has a perpendicular height of 5 = 3 cm, since the height of the rectangular prism is cm. Thus, the triangular face has area 3 3 = 9 cm. Since the length of the triangular prism is 4 cm, then its volume is 9 4 = 36 = 18 cm3. The volume of the combined structure is equal to the sum of the volumes of the two prisms, or = 66 cm 3. Answer: (E) 1. Steve counts forward by 3 beginning at 7. That is, the numbers that Steve counts are each 7 more than some multiple of 3. We can check the given answers to see if they satisfy this requirement by subtracting 7 from each of them and then determining if the resulting number is divisible by 3. We summarize the results in the table below. Answers Result after subtracting 7 Divisible by 3? Yes Yes Yes No No Of the possible answers, Steve only counted 1009, 1006 and Dave counts backward by 5 beginning at 011. That is, the numbers that Dave counts are each some multiple of 5 less than 011. We can check the given answers to see if they satisfy this requirement by subtracting each of them from 011 and then determining if the resulting number is divisible by 5. We summarize the results in the table below. Answers Result after being subtracted from 011 Divisible by 5? No Yes No Yes Yes

76 011 Gauss Contest s Page 7 Of the possible answers, Dave only counted 1006, 1001 and Thus while counting, the only answer that both Steve and Dave will list is 1006 Ȧnswer: (B). In the first 0 minutes, Sheila fills the pool at a rate of 0 L/min and thus adds 0 0 = 400 L of water to the pool. At this time, the pool needs = 3600 L of water to be full. After filling for 0 minutes, water begins to leak out of the pool at a rate of L/min. Since water is still entering the pool at a rate of 0 L/min, then the net result is that the pool is filling at a rate of 0 = 18 L/min. Since the pool needs 3600 L of water to be full and is filling at a rate of 18 L/min, then it will take an additional = 00 minutes before the pool is full of water. Thus, the total time needed to fill the pool is 0+00 = 0 minutes or 3 hours and 40 minutes. Answer: (B) 3. The sum of the units column is E + E + E = 3E. Since E is a single digit, and 3E ends in a 1, then the only possibility is E = 7. Then 3E = 3 7 = 1, and thus is carried to the tens column. The sum of the tens column becomes + B + C + D. The sum of the hundreds column is A + A + A = 3A plus any carry from the tens column. Thus, 3A plus the carry from the tens column is equal to 0. If there is no carry from the tens column, then 3A = 0. This is not possible since A is a single digit positive integer. If the carry from the tens column is 1, then 3A + 1 = 0 or 3A = 19. Again, this is not possible since A is a single digit positive integer. If the carry from the tens column is, then 3A + = 0 or 3A = 18 and A = 6. Since B, C, and D are single digits (ie. they are each less than or equal to 9), then it is not possible for the carry from the tens column to be greater than. Therefore, A = 6 is the only possibility. Since the carry from the tens column is, then the sum of the tens column, + B + C + D, must equal 1. Thus, + B + C + D = 1 or B + C + D = 19. Since A = 6 and E = 7, then the sum A + B + C + D + E = = 3. Note that although we don t know B, C and D, it is only necessary that their sum be 19. Although there are many possibilities, the example with B =, C = 8 and D = 9 is shown Answer: (C) 4. First we recognize that given the conditions for the three selected squares, there are only 6 possible shapes that may be chosen. These are shown below.

77 011 Gauss Contest s Page 8 To determine the number of ways that three squares can be selected, we count the number of ways in which each of these 6 shapes can be chosen from the given figure. The results are summarized in the table below. Shape Number of Each Thus, three of the nine squares can be selected as described in = 19 ways. Answer: (A) 5. We first note that each circle can intersect any other circle a maximum of two times. To begin, the first circle is drawn. The second circle is then drawn overlapping the first, and two points of intersection are created. Since each pair of circles overlap (but are not exactly on top of one another), then the third circle drawn can intersect the first circle twice and the second circle twice. We continue in this manner with each new circle drawn intersecting each of the previously drawn circles exactly twice. That is, the third circle drawn intersects each of the two previous circles twice, the fourth circle intersects each of the three previous circles twice, and so on. Diagrams showing possible arrangements for 3, 4 and 5 circles, each giving the maximum number of intersections, are shown below. The resulting numbers of intersections are summarized in the table below. Circle number drawn Number of new intersections Total number of intersections = = = = = = = = Thus, the greatest possible total number of intersection points using ten circles is = 90.

78 011 Gauss Contest s Page 9 To be complete, we technically need to show that this number is possible, though we don t expect students to do this to answer the question. The diagram below demonstrates a possible positioning of the ten circles that achieves the maximum 90 points of intersection. That is, every pair of circles intersects exactly twice and all points of intersection are distinct from one another. It is interesting to note that this diagram is constructed by positioning each of the ten circles centres at one of the ten vertices of a suitably sized regular decagon, as shown. Answer: (D)

79 011 Gauss Contest s Page 10 Grade 8 1. The fractions 8 and are equivalent fractions. 1 3 To reduce the first to the second, the denominator 1 has been divided by a factor of 4. Therefore, we divide the numerator 8 by the same factor, 4. 8 Thus, = and the value represented by is. 1 3 Answer: (D). Since ground beef sells for $5.00 per kg, then the cost of 1 kg is $ = $ Answer: (C) 3. When the unknown angle y is added to the 90 angle, the result is a complete rotation, or 360. Thus, y + 90 = 360 or y = = 70. Answer: (E) 4. 1 To compare the five fractions we rewrite them each with a common denominator of 100. Then the list equivalent to the given list { 3, 9, 1, 7, } { is 30, 45, 48, 54, }. The largest number in this new list is , and therefore 7 50 is the largest number in the given list. We notice that with the exception of 7, each numerator in the given list is less than one half 50 of its corresponding denominator. Thus, each fraction except 7 has a value that is less than one half. 50 The fraction 7 is larger than one half. 50 Therefore 7 is the largest number in the list. 50 Answer: (D) 5. Since 3 of the 15 balls are red, then the probability that Alex randomly selects a red ball is 3 15 or 1 5. Answer: (A) 6. 1 Since double the original number plus 3 is 3, then double the original number must equal 0 (that is, 3 3). Therefore, the original number is 0 divided by, or 10. Let the original number be represented by the variable x. Then doubling the original number and adding 3 gives x + 3. Thus, x + 3 = 3 or x = 3 3 = 0, so x = 0 = 10. Answer: (B) 7. 1 To make half of the recipe, only half of the 4 1 cups of flour are needed. Since half of 4 is and half of 1 is 1, then 1 cups of flour are needed. 4 4 To make half of the recipe, only half of the 4 1 cups of flour are needed. To determine half of 4 1, we divide 4 1 by, or multiply by 1. Thus, = 9 1 = 9 = 1 cups of flour are needed to make half of the recipe. 4 4 Answer: (B)

80 011 Gauss Contest s Page Since P QR = P RQ, then P QR is an isosceles triangle and P Q = P R = 7. Therefore, the perimeter of P QR is P Q + QR + P R = = 19. Answer: (E) 9. If 15 of 7 students in the class are girls, then the remaining 7 15 = 1 students are boys. The ratio of boys to girls in the class is 1 : 15 = 4 : 5. Answer: (A) 10. Since Kayla ate less than Max and Chris ate more than Max, then Kayla ate less than Max who ate less than Chris. Brandon and Tanya both ate less than Kayla. Therefore, Max ate the second most. Answer: (D) 11. Evaluating each of the expressions, (A): ( 3) = 6 = 36 (B): 3 + = = 7 (C): 3 1 = 8 1 = 7 (D): 3 = 9 4 = 5 (E): (3 + ) = 5 = 5, we see that only expression (D) is equal to 5. Answer: (D) 1. Nick charges $10 per hour of babysitting. If Nick babysits for y hours, then his charge just for babysitting is 10y dollars. In addition, Nick charges a one-time fee of $7 for travel costs. Thus, the expression that represents the total number of dollars that Nick charges for y hours of babysitting is 10y + 7. Answer: (A) 13. Measured in cm, the area of Kalob s window is Measured in cm, twice the area of Kalob s window is which is equal to Thus, a window with dimensions 50 cm 160 cm is a window with area double the area of Kalob s window. Answer: (C) 14. First recognize that the day and the month must be equal. Next, since 3 = 9 and 10 = 100, both the day and the month must be larger than 3 but less than 10 so that the year lies between 01 and 099. We list all possible square root days in the table below. Day and Month Last Two Digits of the Year Date 4 4 = 16 4/4/ = 5 5/5/ = 36 6/6/ = 49 7/7/ = 64 8/8/ = 81 9/9/081 Since these are all actual dates between January 1, 01 and December 31, 099, then the number of square root days is 6. Answer: (E)

81 011 Gauss Contest s Page In CDE, CE = 5, DE = 3, and CDE = 90. By the Pythagorean Theorem, CE = CD + DE or CD = CE DE or CD = 5 3 = 5 9 = 16, so CD = 4 (since CD > 0). In ABC, AB = 9, BC = BD CD = 16 4 = 1, and ABC = 90. By the Pythagorean Theorem, AC = AB + BC or AC = = = 5, so AC = 15 (since AC > 0). Answer: (C) Beatrix is twice the height of Violet who is the height of Georgia. 3 Therefore, Beatrix is times the height of Georgia, or 4 the height of Georgia. 3 3 Let the heights of Beatrix, Violet and Georgia be represented by B, V and G respectively. Since Beatrix is twice the height of Violet, then B = V. Since Violet is the height of Georgia, then V = G. 3 3 Substituting G for V in the first equation, we get B = V = ( G) = 4G Thus, Beatrix s height is 4 of Georgia s height. 3 Answer: (C) 17. Since x can be any value in between 0 and 1, we choose a specific value for x in this range. For example, we allow x to be 1 and then evaluate each of the five expressions. 4 We get, x = 1; 4 x = ( 1 4 ) = 1 ; x = ( ) = = 1; 1 x = = 1; 1 = x 1 = 4. 4 Since 1 is the smallest value, then for any value of x between 0 and 1, 16 x will produce the smallest value of the five given expressions. In fact, no matter what x between 0 and 1 is chosen, x is always smallest. Answer: (B) 18. Assume that each square has side length, and thus has area 4. In square ABCD, diagonal AC divides the square into equal areas. Thus, the area of ACD is one half of the area of square ABCD or. Since AC is the diagonal of square ABCD, ACD = ACB = 90 = 45. Also, DCH = BCH DCB = = 90. Since ACJ is a straight line segment, ACJ = ACD + DCH + HCJ = 180. Thus, HCJ = 180 ACD DCH = = 45. In CHJ, HCJ = 45 and CHJ = 90, so HJC = = 45. Therefore CHJ is isosceles, so CH = HJ = 1, since J is the midpoint of GH. Therefore CHJ has area or 1. Since ACD has area, and CHJ has area 1, then the combined areas of the two shaded regions is + 1 or 5. Since the total area of the two large squares is 8, the fraction of the two squares that is shaded is 5 8 = 5 1 = Answer: (D) 19. We must consider that the integers created could be one-digit, two-digit or three-digit integers. First, consider one-digit integers. Since 1, and 3 are the only digits that may be used, then there are only 3 one-digit positive integers less than 400, namely 1, and 3. Next, consider the number of two-digit integers that can be created. Since digits may be repeated, the integers 11, 1, 13, 1,, 3, 31, 3, 33 are the only possibilities.

82 011 Gauss Contest s Page 13 Thus, there are 9 two-digit positive integers that can be created. Instead of listing these 9 integers, another way to count how many there are is to consider that there are 3 choices for the first digit (either 1, or 3 may be used) and also 3 choices for the second digit (since repetition of digits is allowed), so 3 3 = 9 possibilities. Finally, we count the number of three-digit positive integers. There are 3 choices for the first digit, 3 choices for the second digit and 3 choices for the third digit. Thus, there are = 7 possible three-digit positive integers that can be created. (Note that all of these are less than 400, since the largest of them is 333.) In total, the number of positive integers less than 400 that can be created using only the digits 1, or 3 (with repetition allowed), is = 39. Answer: (D) 0. Since the average height of all students is 103 cm, then the sum of the heights of all students in the class is 103 = 66 cm. Since the average height of the 1 boys in the class is 108 cm, then the sum of the heights of all boys in the class is = 196 cm. The sum of the heights of all the girls in class is the combined height of all students in the class less the height of all the boys in the class, or = 970 cm. Since there are 10 girls in the class, their average height is = 97 cm. Answer: (B) 1. We first consider the minimum number of coins needed to create 99, the greatest amount that we are required to create. We can create 75 with a minimum number of coins by placing three quarters in the collection. Once at 75, we need = 4 more to reach 99. To create 4 with a minimum number of coins we use two dimes and 4 pennies. Therefore, the minimum number of coins required to create 99 is 3 quarters, dimes and 4 pennies, or 9 total coins in the collection. In fact, this is the only group of 9 coins that gives exactly 99. To see this, consider that we must have the 4 pennies and the 3 quarters, but these 7 coins only give 79. We now need exactly 0 and have only two coins left to include in our collection. Therefore the two remaining coins must both be dimes, and thus the only group of 9 coins that gives exactly 99 is 3 quarters, dimes and 4 pennies. We now attempt to check if all other amounts of money less than one dollar can be created using only these 9 coins. Using the 4 pennies, we can create each of the amounts 1,, 3, and 4. However, we don t have a way to create 5 using 3 quarters, dimes and 4 pennies. Thus, at least one more coin is needed. It is clear that if the 10th coin that we add to the collection is a penny, then we will have 5 pennies and thus be able to create 5. However, with 3 quarters, dimes and 5 pennies, we won t be able to create 9. If instead of a penny we add a nickel as our 10th coin in the collection, then we will have 3 quarters, dimes, 1 nickel and 4 pennies. Obviously then we can create 5 using the 1 nickel. In fact, we are now able to create the following sums of money: any amount from 1 to 4 using 4 pennies;

83 011 Gauss Contest s Page 14 any amount from 5 to 9 using 1 nickel and 4 pennies; any amount from 10 to 14 using 1 dime and 4 pennies; any amount from 15 to 19 using 1 dime, 1 nickel and 4 pennies; any amount from 0 to 4 using dimes and 4 pennies. If in addition to the amounts from 1 to 4 we use a quarter, we will be able to create any amount from 5 to 49. Similarly, using quarters we can create any amount from 50 to 74. Finally, using 3 quarters we can create any amount from 75 to 99. Therefore, the smallest number of coins needed to create any amount of money less than one dollar is 10. Can you verify that there is a different combination of 10 coins with which we can create any amount of money less than one dollar? Answer: (A). Consider the possible ways that the numbers from 1 to 9 can be used three at a time to sum to 18. Since we may not repeat any digit in the sum, the possibilities are: , , , , , , Next, consider the row 1 + d + f in Fig.1. Since the sum of every row is 18, d + f = 17. This gives that either d = 8 and f = 9 or d = 9 and f = 8. Next, consider the 3 rows in which x appears. The sums of these 3 rows are a + x + d, b + x + f, and c + x + 6. That is, x appears in exactly 3 distinct sums. Searching our list of possible sums above, we observe that only the numbers 6, 7 and 8 appear in 3 distinct sums. That is, x must equal either 6, 7 or 8. However, the number 6 already appears in the table. Thus, x is not 6. Similarly, we already concluded that either d or f must equal 8. Thus, x is not 8. Therefore, x = 7 is the only possibility. Fig. shows the completed table and verifies that the number represented by x is indeed 7. a b c 1 x d 6 f e Fig Fig. Answer: (C) 3. We first label the trapezoid ABCD as shown in the diagram below. Since AD is the perpendicular height of the trapezoid, then AB and DC are parallel. The area of the trapezoid is AD (AB + DC) or 1 (AB + 16) or 6 (AB + 16). Since the area of the trapezoid is 16, then 6 (AB + 16) = 16 and AB + 16 = 16 or 6 AB + 16 = 7, so AB = 11. Construct a perpendicular from B to E on DC. Since AB is parallel to DE and both AD and BE are perpendicular to DE, then ABED is a rectangle. Thus, DE = AB = 11, BE = AD = 1, and EC = DC DE = = 5. Since BEC = 90, then BEC is a right-angled triangle. Thus by the Pythagorean Theorem, BC = BE + EC or BC = or BC = 169 so

84 011 Gauss Contest s Page 15 BC = 13 (since BC > 0). The perimeter of the trapezoid is AB + BC + CD + DA = = 5 cm. A B 1 cm D E 16 cm C Answer: (B) 4. When Ada glues cube faces together, they must coincide. Also, each of the 4 cubes must have a face that coincides with a face of at least one of the other 3 cubes. We this in mind, we begin by repositioning the 4 identical cubes, attempting to construct figures different from those already constructed. Figure 1 shown below was given in the question. The next 4 figures (numbered to 5) have the same thickness as figure 1. That is, each of the first 5 figures is 1 cube thick from front to back. These are the only 5 figures that can be constructed having this 1 cube thickness and these 5 figures are all unique. Can you verify this for yourself? Attempt to construct a different figure with a 1 cube thickness. Now rotate this figure to see if it will match one of the first 5 figures shown. Next, we consider constructing figures with a front to back depth of cubes. The only 3 unique figures that can be constructed in this way are labeled 6 to 8 below. While figures 7 and 8 look to be the same, there is no way to rotate one of them so that it is identical to the other. Thus, not only are these 3 figures the only figures that can be constructed with a cube thickness, but they are all different from one another. Since no rotation of any of these 3 figures will result in it having a 1 cube thickness from front to back, the figures 6 to 8 are all different than any of the previous 5 figures. With some rotation, we can verify that the only figures having a 3 cube thickness have already been constructed (figures 1, and 5). Similarly, the only figure that can possibly be built with a 4 cube thickness has already been constructed (figure 4). The 8 figures shown below are the only figures that Ada can construct Answer: (D)

85 011 Gauss Contest s Page The given sequence allows for many different patterns to be discovered depending on how terms in the sequence are grouped and then added. One possibility is to add groups of four consecutive terms in the sequence. That is, consider finding the sum of the sequence, S, in the manner shown below. S = (1 + ( 4) + ( 9) + 16) + (5 + ( 36) + ( 49) + 64) + (81 + ( 100) + ( 11) + 144) +... The pattern that appears when grouping terms in this way is that each consecutive group of 4 terms, beginning at the first term, adds to 4. That is, 1+( 4)+( 9)+16 = 4, 5+( 36)+( 49)+64 = 4, 81+( 100)+( 11)+144 = 4, and so on. For now, we will assume that this pattern of four consecutive terms adding to 4 continues and wait to verify this at the end of the solution. Since each consecutive group of four terms adds to 4, the first eight terms add to 8, the first twelve terms add to 1 and the first n terms add to n provided that n is a multiple of 4. Thus, the sum of the first 01 terms is 01, since 01 is a multiple of 4. Since we are required to find the sum of the first 011 terms, we must subtract the value of the 01 th term from our total of 01. We know that the n th term in the sequence is either n or it is n. Therefore, we must determine if the 01 th term is positive or negative. By the alternating pattern of the signs, the first and fourth terms in each of the consecutive groupings will be positive, while the second and third terms are negative. Since the 01 th term is fourth in its group of four, its sign is positive. Thus, the 01 th term is 01. Therefore, the sum of the first 011 terms is the sum of the first 01 terms, which is 01, less the 01 th term, which is 01. Thus, S = = = Verifying the Pattern While we do not expect that students will verify this pattern in the context of a multiple choice contest, it is always important to verify patterns. One way to verify that the sum of each group of four consecutive terms (beginning with the first term) adds to 4, is to use algebra. If the first of four consecutive integers is n, then the next three integers in order are n + 1, n +, and n + 3. Since the terms in our sequence are the squares of consecutive integers, we let n represent the first term in the group of four. The square of the next integer larger than n is (n + 1), and thus the remaining two terms in the group are (n + ) and (n + 3). Since the first and fourth terms are positive, while the second and third terms are negative, the sum of the four terms is n (n + 1) (n + ) + (n + 3). (For example, if n = 5 then we have ) To simplify this expression, we must first understand how to simplify its individual parts such as (n + 1). The expression (n + 1) means (n + 1) (n + 1). To simplify this product, we multiply the n in the first set of brackets by each term in the second set of brackets and do the same for the 1 appearing in the first set of brackets. The operation between each product remains as it appears in the expression, as an addition.

86 011 Gauss Contest s Page 17 That is, (n + 1) = (n + 1) (n + 1) = n n + n n = n + n + n + 1 = n + n + 1 Applying this process again, (n + ) = (n + ) (n + ) = n n + n + n + = n + n + n + 4 = n + 4n + 4, and (n + 3) = (n + 3) (n + 3) = n n + n n = n + 3n + 3n + 9 = n + 6n + 9. Therefore, the sum of each group of four consecutive terms (beginning with the first term) is, n (n + 1) (n + ) + (n + 3) = n (n + n + 1) (n + 4n + 4) + (n + 6n + 9) = n n n 1 n 4n 4 + n + 6n + 9 = n n n + n n 4n + 6n = 4 Answer: (E)

87 Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 010 Gauss Contests (Grades 7 and 8) Wednesday, May 1, 010 s c 009 Centre for Education in Mathematics and Computing

88 010 Gauss Contest s Page 3 Grade 7 1. Reading the number on the vertical axis corresponding to the pet fish, we find that 40 students chose fish as their favourite pet. Answer: (D). By dividing, we find the fraction 0 is equivalent to the decimal We convert this to a percent by multiplying by 100%. Thus, Tanya scored % = 80% on her math quiz. 3. Using the correct order of operations, = = 40. Answer: (C) Answer: (E) 4. To find the location of the point (, 3), we begin at the origin, (0, 0), and move left units and down 3 units. The point (, 3) is located at D. Answer: (D) 5. Going down floors from the 11th floor brings Chaz to the 9th floor. Going down 4 floors from the 9th floor brings Chaz to the 5th floor. Thus, Chaz gets off the elevator on the 5th floor. Answer: (D) 6. The answer, , is 1000 times bigger than This can be determined either by dividing by or by recognizing that the decimal point in is moved three places to the right to obtain Thus, the number that should replace the is Answer: (B) 7. The four angles shown, 150, 90, x, and 90, form a complete rotation, a 360 angle. Thus, x + 90 = 360, or x = = 30. Answer: (D) 8. 1 To build the solid rectangular prism, we could first construct the 4 cm by 3 cm base using 4 3 = 1 of the 1 cm 1 cm 1 cm blocks. Two more layers identical to the first layer, placed on top of the first layer, would give the prism its required 3 cm height. This would require 1 more 1 cm 1 cm 1 cm blocks in layer two and 1 more in layer three, or 1 3 = 36 blocks in total. Equivalently, this question is asking for the volume of the rectangular prism. The volume of a prism is the area of the base times the height, or V = 4 cm 3 cm 3 cm = 36 cm 3. Since the volume of each of the 1 cm 1 cm 1 cm blocks is 1 cm 3, then 36 blocks are needed to build the solid rectangular prism. (The prism can actually be built with 36 blocks as seen in 1.) 4 cm 3 cm Answer: (E)

89 010 Gauss Contest s Page 4 9. If the time reads 3:33, the next time that all of the digits on the clock are equal to one another is 4:44. Since the amount of time between 3:33 and 4:44 is 1 hour and 11 minutes, the shortest length of time in minutes is = 71. Answer: (A) 10. Since 700 is the product of 35 and y, then 35 y = 700 or y = = 0. Since 0 is the product of 5 and x, then 5 x = 0 or x = 0 5 = 4. Answer: (B) We divide the shape into two rectangles, A and B, by constructing the dotted line segment of length units shown. The area of rectangle A is 3 = 6 square units. The length of rectangle B is 6 units plus the length of the dotted line segment, or 6 + = 8. Thus, the area of rectangle B is 8 5 = 40 square units. The area of the entire figure is the sum of the areas of rectangles A and B, or = 46 square units. A 3 B 6 5 By constructing the dotted lines shown, we form a rectangle with length + 6 = 8 units and width = 8 units (in fact, this large rectangle is a square). We find the required area by subtracting the area of rectangle M from the area of the 8 by 8 square. Thus, the area is (8 8) (6 3) = = 46 square units. 3 8 M Answer: (C) 1. If 4 schools each recycle 3 of a tonne of paper, then combined, they recycle = 1 = 3 tonnes of paper. 4 4 Since recycling 1 tonne of paper will save 4 trees, recycling 3 tonnes of paper will save 3 4 = 7 trees. Answer: (B) The mean of 5 consecutive integers is equal to the number in the middle. Since the numbers have a mean of 1, if we were to distribute the quantities equally, we would have 1, 1, 1, 1, and 1. Since the numbers are consecutive, the second number is 1 less than the 1 in the middle, while the fourth number is 1 more than the 1 in the middle. Similarly, the first number is less than the 1 in the middle, while the fifth number is more than the 1 in the middle. Thus, the numbers are 1, 1 1, 1, 1 + 1, 1 +. The smallest of 5 consecutive integers having a mean of 1, is 19. Since 1 is the mean of five consecutive integers, the smallest of these five integers must be less than 1. Suppose that the smallest number is 0.

90 010 Gauss Contest s Page The mean of 0, 1,, 3, and 4 is =. 5 This mean of is greater than the required mean of 1; thus, the smallest of the 5 consecutive integers must be less than 0. Suppose that the smallest number is The mean of 19, 0, 1,, and 3, is = 1, as required. 5 Thus, the smallest of the 5 consecutive integers is 19. Answer: (E) Since the bag contains green mints and red mints only, the remaining 100% 75% = 5% of the mints must be red. Thus, the ratio of the number of green mints to the number of red mints is 75 : 5 = 3 : 1. Since 75% of the mints are green, then 3 of the mints are green. 4 Since the bag contains only green mints and red mints, then 1 3 = 1 of the mints in the bag 4 4 are red. Thus, there are 3 times as many green mints as red mints. The ratio of the number of green mints to the number of red mints is 3 : 1. Answer: (B) 15. The area of square N is four times the area of square M or cm = 400 cm. Thus, each side of square N has length 400 = 0 cm. The perimeter of square N is 4 0 cm = 80 cm. Answer: (C) 16. First we must find the magic constant, that is, the sum of each row, column and diagonal. From column one, we find that the magic constant is (+1) + ( 4) + ( 3) = 6. In the diagonal extending from the top left corner to the bottom right corner, the two existing numbers +1 and 5 have a sum of 4. Thus, to obtain the magic constant of 6 in this diagonal, must occupy the centre square. In the diagonal extending from the bottom left corner to the top right corner, the two numbers 3 and, have a sum of 5. Thus, to obtain the magic constant of 6 in this diagonal, Y must equal 1. The completed magic square is shown below Answer: (A) 17. The smallest possible three-digit integer that is 17 more than a two-digit integer is 100 (100 is 17 more than 83 and 100 is in fact the smallest possible three-digit integer). Notice that 101 is 17 more than 84, 10 is 17 more than 85, and so on. This continues until we reach 117 which is 17 more than 100, but 100 is not a two-digit integer. Thus, 116 is the largest possible three-digit integer that is 17 more than a two-digit integer (116 is 17 more than 99). Therefore, all of the integers from 100 to 116 inclusive, or 17 three-digit integers, are exactly 17 more than a two-digit integer. Answer: (A)

91 010 Gauss Contest s Page We label the 6 points A through F as shown and proceed to connect the points in all possible ways. From point A, 5 line segments are drawn, 1 to each of the other points, B through F. From point B, 4 new line segments are drawn, 1 to each of the points C through F, since the segment AB has already been drawn. This continues, with 3 line segments drawn from point C, from point D, 1 from point E, and 0 from point F since it will have already been joined to each of the other points. In total, there are = 15 line segments. E F D A C B Label the 6 points A through F as shown above. From each of the 6 points, 5 line segments can be drawn leaving the point, 1 to each of the other 5 points. Thus, the total number of line segments leaving the 6 points is 6 5 = 30. However, this counts each of the line segments twice, since each segment will be counted as leaving both of its ends. For example, the segment leaving point A and ending at point D is also counted as a segment leaving point D and ending at point A. Thus, the actual number of line segments is 30 = 15. Answer: (D) 19. The value of any positive fraction is increased by increasing the numerator and/or decreasing the denominator. Thus, to obtain the largest possible sum, we choose 6 and 7 as the numerators, and 3 and 4 as the denominators. We then calculate: = = 46 1 = 3 6 We recognize that 3 is the largest of the 5 possible answers, and thus is the correct response. 6 (This means that we do not need to try ) 4 3 Answer: (E) 0. 1 To determine who cannot be sitting in the middle seat, we may eliminate the 4 people who can be sitting in the middle seat. First, assume that Sally and Mike, who must be beside one another, are in seats 1 and, or in seats and 1. Since Andy and Jen are not beside each other, either Andy is in seat 3 (the middle seat) and Jen is in seat 5, or vice versa. Thus, Andy and Jen can each be sitting in the middle seat and are eliminated as possible choices. Next, assume that Sally and Mike are in seats and 3, or in seats 3 and. That is, either Sally is in the middle (seat 3), or Mike is. In either case, seats 1, 4 and 5 are empty, allowing either Andy or Jen to choose seat 1 and hence, they are not next to one another. This demonstrates that Sally and Mike can each be sitting in the middle seat.

92 010 Gauss Contest s Page 7 Having eliminated Andy, Jen, Sally, and Mike, it must be Tom who cannot be sitting in the middle seat. Assume that Tom is sitting in the middle (seat 3). Since Sally and Mike are seated beside each other, they are either sitting in seats 4 and 5 or seats 1 and. In either case, seats 1 and remain empty or seats 4 and 5 remain empty. However, Andy and Jen cannot sit beside each other. Therefore, this arrangement is not possible. Thus, Tom cannot be sitting in the middle seat. Since the question implies that there is a unique answer, then Tom is the answer. Answer: (E) 1. Traveling at a constant speed of 15 km/h, in 3 hours the bicycle will travel 15 3 = 45 km. At the start, the bicycle was 195 km ahead of the bus. Therefore, in order to catch up to the bicycle, the bus must travel 195 km plus the additional 45 km that the bicycle travels, or = 40 km. To do this in 3 hours, the bus must travel at an average speed of 40 3 = 80 km/h. Answer: (B). When tossing a single coin, there are two possible outcomes, a head (H) or a tail (T). When tossing coins, there are = 4 possible outcomes. These are HH, HT, TH, and TT. When tossing 3 coins, there are = 8 possible outcomes. These are HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT. Of these 8 possible outcomes, there are winning outcomes, HHH and TTT. Thus, the probability of winning the Coin Game is 8 = 1 4. Answer: (B) 3. Since M O M = 49, either M = 7 and O = 1 or M = 1 and O = 49. However, since the value of the word TOTE is 18, O cannot have a value of 49 because 18 is not divisible by 49. Thus, M = 7 and O = 1. Since T O T E = 18 and O = 1, we have T T E = 18. Therefore, either T = 3 and E = or T = 1 and E = 18. However, O = 1, and since every letter has a different value, T cannot be equal to 1. Thus, T = 3 and E =. The value of the word TEAM is 168, so T E A M = 168, or 3 A 7 = 168. Thus, 4 A = 168 or A = = 4. The value of the word HOME is 70, so H O M E = 70, or H 1 7 = 70. Thus, 14 H = 70 or H = = 5. Finally, the value of the word MATH is M A T H = = 40. Answer: (C)

93 010 Gauss Contest s Page 8 4. The sum of two even numbers is even. The sum of two odd numbers is even. The sum of an odd number and an even number is odd. Thus, for the sum m + n to be even, both m and n must be even, or they must both be odd. If m =, then n must be even and greater than. Thus, if m = then n can be 4, 6, 8, 10, 1, 14, 16, 18, or 0. This gives 9 different pairs (m, n) when m =. If m = 4, then n must be even and greater than 4. Thus, if m = 4 then n can be 6, 8, 10, 1, 14, 16, 18, or 0. This gives 8 different pairs (m, n) when m = 4. Continuing in this manner, each time we increase m by, the number of choices for n, and thus for (m, n), decreases by 1. This continues until m = 18, at which point there is only one choice for n, namely n = 0. Therefore, the total number of different pairs (m, n) where both m and n are even is, = 45. Similarly, if m = 1, then n must be odd and greater than 1. Thus, if m = 1, then n can be 3, 5, 7, 9, 11, 13, 15, 17, or 19. This gives 9 different pairs (m, n) when m = 1. If m = 3, then n must be odd and greater than 3. Thus, if m = 3 then n can be 5, 7, 9, 11, 13, 15, 17, or 19. This gives 8 different pairs (m, n) when m = 3. Continuing in this manner, each time we increase m by, the number of choices for n, and thus for (m, n), decreases by 1. This continues until m = 17, at which point there is only one choice for n, namely n = 19. Therefore, the total number of different pairs (m, n) where both m and n are odd is, = 45. Thus, the total number of different pairs (m, n) using numbers from the list {1,, 3,..., 0} such that m < n and m + n is even is = 90. Answer: (B) 5. Together, Hose A and Hose B fill the pool in 6 hours. Thus, it must take Hose A more than 6 hours to fill the pool when used by itself. Therefore, a 7, since a is a positive integer. Similarly, it must take Hose B more than 6 hours to fill the pool when used by itself. Therefore, b 7, since b is a positive integer. When used by itself, the fraction of the pool that Hose A fills in 6 hours is 6 a. When used by itself, the fraction of the pool that Hose B fills in 6 hours is 6 b. When used together, Hose A and Hose B fill the pool once in 6 hours. Thus, = 1. a b Since a 7, b 7, and both a and b are integers, then we can find values for a and b that satisfy the equation = 1 by using systematic trial and error. a b For example, let a = 7. Then = 1, or 6 = 1 6, or 6 = 1. 7 b b 7 b 7 Since 6 = 1, then b = 4 and a = 7 is one possible solution to the equation = a b Compare this to what happens when we let a = 11. We have = 1, or 6 = 1 6, or 6 = 5, or 5b = b b 11 b 11 Since there is no integer value for b that makes 5b equivalent to 66, then a = 11 does not give a possible solution. The possible solutions found by systematic trial and error are shown below. a b

94 010 Gauss Contest s Page 9 Any value for a larger than 4 requires b to be smaller than 7, but we know that b 7. Thus, there are only 9 different possible values for a. Note: We can reduce the time it takes to complete this trial and error above by recognizing that in the equation = 1, the a and b are interchangeable. a b That is, interchanging a and b in the equation, gives = 1, which is the same equation. b a For example, this tells us that since a = 7, b = 4 satisfies the equation, then a = 4, b = 7 satisfies the equation as well. Moreover, if the pair (a, b) satisfies the equation, then (b, a) satisfies the equation, and if (a, b) does not satisfy the equation, then (b, a) does not satisfy the equation. This interchangeability of a and b is seen in the symmetry of the list of possible solutions above. Recognizing that this symmetry must exist allows us to quickly determine the 4 remaining solutions that follow after a = 1, b = 1. Answer: (C)

95 010 Gauss Contest s Page 10 Grade 8 1. Using the correct order of operations, = = 4.. The athlete who won the race is the one who had the shortest running time. Thus, athlete C won the race. Answer: (A) Answer: (C) 3. Substituting x = and y = 1 into the expression x 3y, we have 3 1. Using the correct order of operations, 3 1 = 4 3 = 1. Answer: (B) 4. 1 Evaluating the left side of the equation, we get 44 5 = Thus, the number that should replace the is = 11. The left side of the equation, 44 5, can be rewritten as Since = , then we can write = 100, so the number that should replace the is 11. Answer: (A) 5. The area of the rectangle, 1, is found by multiplying its length by its width. Since the length and width must be whole numbers, the only possible dimensions are: 1 by 1, 6 by, and 4 by 3. (We recognize that 1 by 1, by 6, and 3 by 4 are also possibilities, but the perimeter of the rectangle is unchanged by reversing the dimensions.) The perimeter of a rectangle is found by doubling the sum of the length and width. The results are shown in the table below. Length Width Perimeter Therefore, the smallest possible perimeter of a rectangle with the given conditions is 14. Answer: (D) 6. We first recognize that 1 is a common fraction in each of the five sums, and so the relative size 4 of the sums depends only on the other fractions. Since 1 is the largest of the fractions 1, 1, 1, 1, 1, we conclude that is the largest sum Answer: (C) 7. Since 1 gram is the approximate weight of 15 seeds, 300 grams is the approximate weight of = 4500 seeds. Therefore, there are approximately 4500 seeds in the container. Answer: (B) 8. The first time after 10:5 at which all of the digits on the clock will be equal to one another is 11:11. Thus, the shortest length of time required is the elapsed time between 10:5 and 11:11, or 46 minutes (because the time from 10:5 to 11:00 is 35 minutes and the time from 11:00 to 11:11 is 11 minutes). Answer: (D)

96 010 Gauss Contest s Page Chris was given 1 3 of 84 cookies, or = 84 3 = 8 cookies. He ate 3 4 of the 8 cookies he was given, or = 84 4 = 1 cookies. Answer: (E) In ABC shown below, BAC = 180 ABC ACB = = 30. Since ADC is a straight angle, ADE = 180 CDE = = 13. In AED, AED = 180 ADE EAD = = 18. Since AEB is a straight angle, DEB = 180 AED = = 16. Thus, the value of x is 16. A E x 48 D F 60 B The sum of the interior angles of a quadrilateral is 360. In quadrilateral BCDE shown above, DEB = 360 EDC DCB CBE = = 16. C Thus, the value of x is 16. Answer: (E) The mean of 5 consecutive integers is equal to the number in the middle. Since the numbers have a mean of 1, if we were to distribute the quantities equally, we would have 1, 1, 1, 1, and 1. Since the numbers are consecutive, the second number is 1 less than the 1 in the middle, while the fourth number is 1 more than the 1 in the middle. Similarly, the first number is less than the 1 in the middle, while the fifth number is more than the 1 in the middle. Thus, the numbers are 1, 1 1, 1, 1 + 1, 1 +. The smallest of 5 consecutive integers having a mean of 1, is 19. Since 1 is the mean of five consecutive integers, the smallest of these five integers must be less than 1. Suppose the smallest integer is The mean of 0, 1,, 3, and 4 is =. 5 This mean of is greater than the required mean of 1; thus, the smallest of the 5 consecutive integers must be less than 0. Suppose the smallest integer is The mean of 19, 0, 1,, and 3 is = 1, as required. 5 Thus, the smallest of 5 consecutive integers having a mean of 1, is 19. Answer: (E)

97 010 Gauss Contest s Page 1 1. For every 3 white balls in the jar, there are red balls in the jar. Since there are 9 white balls in the jar, which is 3 groups of 3 white balls, there must be 3 groups of red balls in the jar. Thus, there are 3 = 6 red balls in the jar. Answer: (D) 13. Evaluating, ( 11 1 Since > = 1 ) = ( 11 1 ) ( 11 ) 1 = = and < = 1, the value of ( 11 1 ) is between 1 and 1. Answer: (B) 14. During the 5 games, Gina faced = 65 shots in total. During the 5 games, Gina saved = 5 of the 65 shots. Thus, the percentage of total shots saved is % = % = 80%. Answer: (C) 15. To find the smallest possible sum, we first choose the tens digit of each number to be as small as possible. Therefore, we choose 5 and 6 as the two tens digits. Next, we choose the units digits to be as small as possible. Since 7 and 8 are each less than 9, we choose 7 and 8 as the two units digits. Using 5 and 6 as the tens digits, 7 and 8 as the units digits, we evaluate the only two possibilities (Can you see why these two sums should be equal?) The smallest possible sum is 15. Answer: (B) Since AQ = 0 and AB = 1, then BQ = AQ AB = 0 1 = 8. Thus, P B = P Q BQ = 1 8 = 4. Since P S = 1, the area of rectangle P BCS is 1 4 = 48. A P B Q D S C R The sum of the areas of squares ABCD and P QRS is (1 1) = 144 = 88. The area of rectangle AQRD is 1 0 = 40. The sum of the areas of ABCD and P QRS is equal to the sum of the areas of AP SD, P BCS, P BCS, and BQRC. The area of rectangle AQRD is equal to the sum of the areas of AP SD, P BCS, and BQRC. Therefore, the sum of the areas of ABCD and P QRS, minus the area of AQRD, is the area of P BCS. Thus, the shaded rectangle P BCS has area = 48. Answer: (C)

98 010 Gauss Contest s Page Label the 8 points A through H as shown and proceed to connect the points in all possible ways. From point A, 7 line segments are drawn, 1 to each of the other points, B through H. From point B, 6 new line segments are drawn, 1 to each of the points C through H, since the segment AB has already been drawn. This continues, with 5 line segments drawn from point C, 4 from D, 3 from E, from F, 1 from G and 0 from point H since it will already be joined to each of the other points. In total, there are = 8 line segments. Label the 8 points A through H as shown above and proceed to connect the points in all possible ways. From each of the 8 points, 7 line segments can be drawn leaving the point, 1 to each of the other 7 points. Thus, the total number of line segments leaving the 8 points is 8 7 = 56. However, this counts each of the line segments twice, since each segment will be counted as leaving both of its ends. For example, the segment leaving point A and ending at point D is also counted as a segment leaving point D and ending at point A. Thus, the actual number of line segments is half of 56 or 56 = 8. Answer: (E) 18. Traveling at a constant speed of 15 km/h, in 3 hours the bicycle will travel 15 3 = 45 km. At the start, the bicycle was 195 km ahead of the bus. Therefore, in order to catch up to the bicycle, the bus must travel 195 km plus the additional 45 km that the bicycle travels, or = 40 km. To do this in 3 hours, the bus must travel at an average speed of 40 3 = 80 km/h. Answer: (B) 19. Figure 1 is formed with 1 square. Figure is formed with squares. Figure 3 is formed with = squares. Figure 4 is formed with = squares. Figure 5 is formed with = squares. Thus, the number of groups of 4 squares needed to help form the Figure is increasing by 1. Also, in each case the number of groups of 4 squares needed is one less than the Figure number. For example, Figure 6 will be formed with 5 groups of 4 squares plus 1 additional square. In general, we can say that Figure N will be formed with N 1 groups of 4 squares, plus 1 additional square. Thus, Figure 010 will be formed with = = 8037 squares. Answer: (A) F G E H D A C B 0. Position point T on QR such that P T is perpendicular to QR. Line segment P T is the height of P QS with base QS, and is also the height of P RS with base SR. Since P QS and P RS have equal heights and equal areas, their bases must be equal. Thus, QS = SR. Q P T S R Answer: (D)

99 010 Gauss Contest s Page Since ACE is a straight angle, ACB = = 75. In ABC, BAC = 180 ABC ACB = = 30. Since AB is parallel to DC, ACD = BAC = 30 (alternate angles). In ADC, DAC = 180 ADC ACD = = 35. Thus, the value of x is 35. A x 115 D B C E Answer: (A). We first recognize that in the products, r s, u r and t r, r is the only variable that occurs in all three. Thus, to make r s + u r + t r as large as possible, we choose r = 5, the largest value possible. Since each of s, u and t is multiplied by r once only, and the three products are then added, it does not matter which of s, u or t we let equal, 3 or 4, as the result will be the same. Therefore, let s =, u = 3 and t = 4. Thus, the largest possible value of r s+u r +t r is = = 45. Answer: (B) 3. Since Kevin needs 1 hours to shovel all of his snow, he shovels 1 of his snow every hour. 1 Since Dave needs 8 hours to shovel all of Kevin s snow, he shovels 1 of Kevin s snow every hour. 8 Similarly, John shovels 1 of Kevin s snow every hour, and Allison shovels 1 of Kevin s snow 6 4 every hour. 1 Together, Kevin, Dave, John, and Allison can shovel = = of Kevin s snow every hour. Therefore, together they can shovel = 1 = 15 = 1 of Kevin s snow every minute Thus, by shoveling 1 of Kevin s snow per minute, together they will shovel all of Kevin s snow 96 in 96 minutes. Answer: (D) 4. Label points A and B, the points of intersection of the two circles, and point O, the centre of the left circle. Construct line segment AB, which by symmetry divides the shaded area in half. Construct radii OA and OB with OA = OB = 10 cm. Since each circle contains 5% or 1 of the other circle s circumference, AOB = = Thus, the area of sector AOB is 1 of the area of the entire circle, 4 or 1 4 πr = 1 4 π10 = 5π cm. OA OB The area of AOB is = = 50 cm. The area remaining after AOB is subtracted from sector AOB is equal to half of the shaded area. Thus, the shaded area is (5π 50) (8.5398) = cm. The area of the shaded region is closest to cm. O A B Answer: (A)

100 010 Gauss Contest s Page We are given that the first two terms of a 10 term sequence are 1 and x. Since each term after the second is the sum of the previous two terms, then the third term is 1 + x. Since the fourth term is the sum of the second and third terms, then the fourth term is x + (1 + x) = 1 + x. Continuing in this manner, we construct the 10 term sequence: 1, x, 1 + x, 1 + x, + 3x, 3 + 5x, 5 + 8x, x, x, x. Each of the second through tenth terms is dependent on the value of x, and thus, any one of these terms could potentially equal 463. For the second term to equal 463, we need x = 463, which is possible since the only requirement is that x is a positive integer. Thus, if x = 463 then 463 appears as the second term in the sequence. For the third term to equal 463, we need 1 + x = 463, or x = 46. Thus, if x = 46 then 463 appears as the third term in the sequence. For the fourth term to equal 463, we need 1 + x = 463, or x = 46 or x = 31. Thus, if x = 31 then 463 appears as the fourth term in the sequence. For the fifth term to equal 463, we need + 3x = 463, or 3x = 461 or x = However, 461 is not an integer, and thus, 463 cannot appear as the fifth term in the sequence. 3 We continue in this manner and summarize all the results in the table below. Term Expression Equation Value of x Is x an integer? nd x x = 463 x = 463 Yes 3rd 1 + x 1 + x = 463 x = 46 Yes 4th 1 + x 1 + x = 463 x = 31 Yes 5th + 3x + 3x = 463 x = No 6th 3 + 5x 3 + 5x = 463 x = 9 Yes 7th 5 + 8x 5 + 8x = 463 x = No 8th x x = 463 x = 35 Yes 9th x x = 463 x = No 10th x x = 463 x = 13 Yes Therefore, the sum of all possible integer values of x for which 463 appears in the sequence is = 196. Answer: (B)

101 Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 009 Gauss Contests (Grades 7 and 8) Wednesday, May 13, 009 s c 008 Centre for Education in Mathematics and Computing

102 009 Gauss Contest s Page 3 Grade 7 1. Adding, = = Since the triangle is equilateral, all sides are equal in length. Therefore, the perimeter of the triangle is = 8 3 = 4. Answer: (A) Answer: (C) 3. The numbers 1, 14 and 16 are even, and therefore divisible by so not prime. The number 15 is divisible by 5; therefore, it is also not prime. Each of the remaining numbers, 11, 13 and 17, has no positive divisor other than 1 and itself. Therefore, 3 numbers in the list are prime. Answer: (D) 4. 1 Since each number in the set is between 0 and 1, they can be ordered from smallest to largest by comparing their tenths digits first. In order from smallest to largest the list is The smallest number in the list is {0.05, 0.5, 0.37, 0.40, 0.81}. Consider the equivalent fraction for each decimal: 0.40 = , 0.5 =, 0.37 =, 0.05 = 5 81, and 0.81 = Since the denominators all equal 100, we choose the fraction with the smallest numerator. Therefore, 0.05 = 5 is the smallest number in the set. 100 Answer: (D) 5. The x-coordinate of point P lies between and 0. The y-coordinate lies between and 4. Of the possible choices, ( 1, 3) is the only point that satisfies both of these conditions. Answer: (E) 6. The temperature in Vancouver is C. The temperature in Calgary is C 19 C = 3 C. The temperature in Quebec City is 3 C 11 C = 8 C. Answer: (C) 7. Since a real distance of 60 km is represented by 1 cm on the map, then a real distance of 540 km is represented by cm or 9 cm on the map. Answer: (A) 8. The sum of the three angles in any triangle is always 180. In P QR, the sum of P and Q is 60, and thus R must measure = 10. Answer: (C)

103 009 Gauss Contest s Page 4 9. The first Venn diagram below shows that there are 30 students in the class, 7 students have been to Mexico, 11 students have been to England and 4 students have been to both countries. Of the 7 students that have been to Mexico, 4 have also been to England. Therefore, 7 4 = 3 students have been to Mexico and not England. Of the 11 students that have been to England, 4 have also been to Mexico. Therefore, 11 4 = 7 students have been to England and not Mexico. Class (30) Class (30) M (7) E (11) M (7) E (11) Therefore, 3 students have been to Mexico only, 7 students have been to England only, and 4 students have been to both. In the class of 30 students, this leaves = 16 students who have not been to Mexico or England. Answer: (B) 10. Consider rotating the horizontal line segment F G (as shown below) 180 about point F. A 180 rotation is half of a full rotation. Point F stays fixed, while segment F G rotates to the left of F (as does the rest of the diagram). Figure C shows the correct result. 16 F G G F Answer: (C) 11. Since Scott runs 4 m for every 5 m Chris runs, Scott runs 4 of the distance that Chris runs in 5 the same time. When Chris crosses the finish line he will have run 100 m. When Chris has run 100 m, Scott will have run = 80 m. 5 Answer: (E) 1. The area of a triangle can be calculated using the formula Area = 1 base height. The area is 7 cm and the base measures 6 cm. Substituting these values into the formula, A = 1 b h becomes 7 = 1 6 h or 7 = 3h. Therefore, h = 9 cm. Answer: (A) 13. There are 60 seconds in a minute, 60 minutes in an hour, 4 hours in a day and 7 days in a week. Therefore, the number of seconds in one week is Answer: (D) S represents a value of approximately 1.5 on the number line, while T is approximately 1.6. Then S T is approximately equal to = R is the only value on the number line that is slightly less than 1 and therefore best represents the value of S T. S is slightly less than T, so S T is slightly less than 1. Thus, S T is best represented by R. Answer: (C)

104 009 Gauss Contest s Page For the sum to be a maximum, we try to use the largest divisor possible. Although 144 is the largest divisor, using it would require that the remaining two divisors both equal 1 (since the divisors are integers). Since the question requires the product of three different divisors, 144 = is not possible and the answer cannot be = 146 or (C). The next largest divisor of 144 is 7 and 144 = 7 1. Now the three factors are different and their sum is = 75. Since 75 is the largest possible answer remaining, we have found the maximum. Answer: (B) For the square to have an area of 5, each side length must be 5 = 5. The rectangle s width is equal to that of the square and therefore must also be 5. The length of the rectangle is double its width or 5 = 10. The area of the rectangle is thus 5 10 = 50. The rectangle has the same width as the square but twice the length. Thus, the rectangle s area is twice that of the square or 5 = The six other players on the team averaged 3.5 points each. The total of their points was = 1. Vanessa scored the remainder of the points, or 48 1 = 7 points. Answer: (D) Answer: (E) 18. Since x and z are positive integers and xz = 3, the only possibilities are x = 1 and z = 3 or x = 3 and z = 1. Assuming that x = 1 and z = 3, yz = 6 implies 3y = 6 or y =. Thus, x = 1 and y = and xy =. This contradicts the first equation xy = 18. Therefore, our assumption was incorrect and it must be true that x = 3 and z = 1. Then yz = 6 and z = 1 implies y = 6. Checking, x = 3 and y = 6 also satisfies xy = 18, the first equation. Therefore, the required sum is x + y + z = = 10. Answer: (B) 19. The value of all quarters is $ Each quarter has a value of $0.5. There are thus = 40 quarters in the jar. Similarly, there are = 00 nickels, and = 1000 pennies in the jar. In total, there are = 140 coins in the jar. the number of quarters The probability that the selected coin is a quarter is the total number of coins = = Since V is the midpoint of P R, then P V = V R. Since UV RW is a parallelogram, then V R = UW. Since W is the midpoint of US, then UW = W S. Thus, P V = V R = UW = W S. Answer: (B)

105 009 Gauss Contest s Page 6 Similarly, QW = W R = UV = V T. Also, R is the midpoint of T S and therefore, T R = RS. Thus, V T R is congruent to W RS, and so the two triangles have equal area. Diagonal V W in parallelogram UV RW divides the area of the parallelogram in half. Therefore, UV W and RW V have equal areas. In quadrilateral V RSW, V R = W S and V R is parallel to W S. Thus, V RSW is a parallelogram and the area of RW V is equal to the area of W RS. Therefore, V T R, W RS, RW V, and U V W have equal areas, and so these four triangles divide ST U into quarters. Parallelogram UV RW is made from two of these four quarters of ST U, or one half of ST U. The area of parallelogram UV RW is thus 1 of 1, or 1. Answer: (B) 1. Together, Lara and Ryan ate = = 11 of the pie Therefore, 1 11 = 9 of the pie remained. 0 0 The next day, Cassie ate of the pie that remained. 3 This implies that 1 = 1 of the pie that was remaining was left after Cassie finished eating. 3 3 Thus, 1 of 9, or 3 of the original pie was not eaten Answer: (D). The first row is missing a 4 and a. Since there is already a in the second column (in the second row), the first row, second column must contain a 4 and the first row, fourth column must contain a. To complete the upper left square, the second row, first column must contain a 3. The second row is now missing both a 4 and a 1. But the fourth column already contains a 4 (in the fourth row), therefore the second row, fourth column must contain a 1. To complete the fourth column, we place a 3 in the third row. Now the P cannot be a 3, since there is already a 3 in the third row. Also, the P cannot be a 4 or a, since the second column already contains these numbers. By process of elimination, the digit 1 must replace the P. Answer: (A) 3. 1 We can suppose that the jug contains 1 litre of water at the start. The following table shows the quantity of water poured in each glass and the quantity of water remaining in each glass after each pouring, stopping when the quantity of water remaining is less than 0.5 L. Number of glasses Number of litres poured Number of litres remaining 1 10 % of 1 = = % of 0.9 = = % of 0.81 = = % of 0.79 = = % of = = % of = = % of = = We can see from the table that the minimum number of glasses that Kim must pour so that less than half of the water remains in the jug is 7. Removing 10% of the water from the jug is equivalent to leaving 90% of the water in the jug.

106 009 Gauss Contest s Page 7 Thus, to find the total fraction remaining in the jug after a given pour, we multiply the previous total by 0.9. We make the following table; stopping when the fraction of water remaining in the glass is first less than 0.5 (one half). Number of glasses poured Fraction of water remaining = = = = = = = We can see from the table that the minimum number of glasses that Kim must pour so that less than half of the water remains in the jug is 7. Answer: (C) 4. 1 Draw line segment QR parallel to DC, as in the following diagram. This segment divides square ABCD into two halves. Since triangles ABQ and RQB are congruent, each is half of rectangle ABRQ and therefore one quarter of square ABCD. Draw line segment P S parallel to DA, and draw line segment P R. Triangles P DQ, P SQ, P SR and P CR are congruent. Therefore each is one quarter of rectangle DCRQ and therefore one eighth of square ABCD. A Q S D P C Quadrilateral QBCP therefore represents = 5 of square ABCD. Its area is therefore 5 of the area of the square. 8 Therefore, 5 of the area of the square is equal to 15. Therefore, 1 of the area of the square is 8 8 equal to 3. Therefore the square has an area of 4. B R Draw a line segment from Q to R, the midpoint of BC. Draw a line segment from P to S, the midpoint of QR. Let the area of QSP equal x. Thus, the area of QDP is also x and QDP S has area x. Square SP CR is congruent to square QDP S and thus has area x. Rectangle QDCR has area 4x, as does the congruent rectangle AQRB. Also, AQB and BRQ have equal areas and thus, each area is x.

107 009 Gauss Contest s Page 8 A B Q x x x S x x x R D P C Quadrilateral QBCP is made up of BRQ, QSP and square SP CR, and thus has area x + x + x = 5x. Since quadrilateral QBCP has area 15, then 5x = 15 or x = 3. Therefore, the area of square ABCD, which is made up of quadrilateral QBCP, AQB and QDP, is 5x + x + x = 8x = 8(3) = 4. Answer: (E) 5. Labeling the diagram as shown below, we can describe paths using the points they pass through. E N F D A C M The path MADN is the only path of length 3 (traveling along 3 diagonals). Since the diagram is symmetrical about MN, all other paths will have a reflected path in the line M N and therefore occur in pairs. This observation alone allows us to eliminate (B) and (D) as possible answers since they are even. The following table lists all possible paths from M to N traveling along diagonals only. Path length Path Name Reflected Path 3 MADN same 5 MABCDN MAF EDN 9 M ABCDEF ADN M AF EDCBADN MABCDAF EDN MAF EDABCDN MADCBAF EDN MADEF ABCDN At this point we have listed 9 valid paths. Since paths occur in pairs (with the exception of MADN), the next possible answer would be 11. Since 11 is not given as an answer (and 9 is the largest possible answer given), we can be certain that we have found them all. Answer: (E) B

108 009 Gauss Contest s Page 9 Grade 8 1. Using the correct order of operations, = = 10.. Calculating, 10 + ( 1) =. Answer: (B) Answer: (D) 3. Using a 1 litre jug of water, Jack could fill two 0.5 (one half) litre bottles of water. Using a 3 litre jug of water, Jack could fill 3 = 6 bottles of water. (Check: = 3 1 = 3 = 6) Answer: (C) 4. Since AB is a line segment, ACD+ DCE+ ECB = 180 or 90 + x + 5 = 180 or x = or x = 38. D E A x 5 C B Answer: (B) 5. Calculating, 7 9 = 7 9 = = 0.7. Rounded to decimal places, 7 9 is Answer: (C) 6. The graph shows that vehicle X uses the least amount of fuel for a given distance. Therefore, it is the most fuel efficient vehicle and will travel the farthest using 50 litres of fuel. Answer: (D) 7. Kayla spent 1 1 $100 = $5 on rides and $100 = $10 on food The total that she spent was $35. Answer: (E) 8. The polyhedron has 6 faces and 8 vertices. The equation F + V E = becomes E = or 14 E = or E = 14 = 1. Therefore, the polyhedron has 1 edges. Answer: (A) 9. Eliminating multiple occurrences of the same letter, the word PROBABILITY uses 9 different letters of the alphabet, A, B, I, L, O, P, R, T, and Y. Since there are 6 letters in the alphabet, the probability that Jeff picks one of the 9 different letters in PROBABILITY is 9 6. Answer: (A) Since the two numbers differ by but add to 0, the smaller number must be 1 less than half of 0 while the larger number is 1 greater than half of twenty. The smaller number is 9 and the larger is 11. Since the numbers differ by, let the smaller number be represented by x and the larger number be represented by x +. Since their sum is 0, then x + x + = 0 or x = 18 or x = 9. The smaller number is 9 and the larger is 11. Answer: (A)

109 009 Gauss Contest s Page Since ABC = ACB, then ABC is isosceles and AB = AC. Given that ABC has a perimeter of 3, AB + AC + 1 = 3 or AB + AC = 0. But AB = AC, so AB = 0 or AB = 10. Answer: (C) 1. Substituting C = 10, the equation F = 9C + 3 becomes F = = = A temperature of 10 degrees Celsius is equal to 50 degrees Fahrenheit. Answer: (D) 13. Beginning with the positive integer 1 as a number in the first pair, we get the sum 101 = From this point we can continue to increase the first number by one while decreasing the second number by one, keeping the sum equal to 101. The list of possible sums is: 101 = = = = After this point, the first number will no longer be smaller than the second if we continue to add 1 to the first number and subtract 1 from the second. There are 50 possible sums in all. Answer: (A) 14. The six other players on the team averaged 3.5 points each. The total of their points was = 1. Vanessa scored the remainder of the points, or 48 1 = 7 points.. Answer: (E) 15. Triangle P QR is a right-angled triangle since P QR = 90 (because P QRS is a rectangle). In P QR, the Pythagorean Theorem gives, P R = P Q + QR 13 = 1 + QR 169 = QR = QR QR = 5 So QR = 5 since QR > 0. The area of P QRS is thus 1 5 = 60. Answer: (B) 16. When it is 3:00 p.m. in Victoria, it is 6:00 p.m. in Timmins. Therefore, Victoria time is always 3 hours earlier than Timmins time. When the flight arrived at 4:00 p.m. local Timmins time, the time in Victoria was 1:00 p.m. The plane left at 6:00 a.m. Victoria time and arrived at 1:00 p.m. Victoria time. The flight was 7 hours long. Answer: (D)

110 009 Gauss Contest s Page The value of all quarters is $ Each quarter has a value of $0.5. There are thus = 40 quarters in the jar. Similarly, there are = 00 nickels, and = 1000 pennies in the jar. In total, there are = 140 coins in the jar. the number of quarters The probability that the selected coin is a quarter is the total number of coins = = Answer: (B) 18. Of the 40 students, 1 did not like either dessert. Therefore, 40 1 = 8 students liked at least one of the desserts. But 18 students said they liked apple pie, 15 said they liked chocolate cake, and = 33, so 33 8 = 5 students must have liked both of the desserts. Answer: (E) 19. In the ones column, P + Q ends in 9. So P + Q = 9 or P + Q = 19. Since P is at most 9 and Q is at most 9, then P + Q is at most 18. Therefore, P + Q = 9 since P + Q cannot equal 19. In the tens column, since Q + Q ends in zero, Q + Q equals 0 or 10. Therefore, either Q = 0 or Q = 5. If Q = 0, there would be no carry to the hundreds column where P + Q (plus no carry) ends in a zero. This is not possible since we already determined that P + Q = 9. Therefore, Q = 5 and P = 4, giving a 1 carried from the tens column to the hundreds column. In the hundreds column, we have which gives a 1 carried from the hundreds column to the thousands column. Then R plus the 1 carried equals, so R = 1. Thus, P + Q + R = = 10. P Q P + R Q Q Q Answer: (B) 0. Since the area of the square is 144, each side has length 144 = 1. The length of the string equals the perimeter of the square which is 4 1 = 48. The largest circle that can be formed from this string has a circumference of 48 or πr = 48. Solving for the radius r, we get r = π The maximum area of a circle that can be formed using the string is π( 48 π ) π(7.64) 183. Answer: (E) 1. For the sum to be a maximum, we must choose the three smallest divisors in an effort to make the fourth divisor as large as possible. The smallest 3 divisors of 360 are 1, and 3, making 360 = 60 the fourth divisor. 1 3 We note here that 1, and 3 are the smallest three different divisors of 360. Therefore, it is not possible to use a divisor greater than 60, since there aren t three divisors smaller than 1, and 3. Replacing the divisor 60 with a smaller divisor will decrease the sum of the four divisors. To see this, we recognize that the product of 3 different positive integers is always greater than

111 009 Gauss Contest s Page 1 or equal to the sum of the 3 integers. For example, 1 4 = 8 > = 7. The next largest divisor less than 60 is 45, thus the remaining three divisors would have a product of = 8, and therefore have a sum that is less than or equal to 8. This gives a combined sum that is less than or equal to 45+8 = 53, much less than the previous sum of = 66. In the same way, we obtain sums smaller than 66 if we consider the other divisors of 360 as the largest of the four integers. Therefore, the maximum possible sum is 66. Answer: (B). The first vertical line through the letter S cuts the S into 4 pieces, pieces to the left of the line and to the right. Each additional vertical line adds 3 new pieces while preserving the 4 original pieces. The chart below shows the number of pieces increasing by three with each additional line drawn. Lines Pieces We want 154 pieces. Since the first line gives 4 pieces, we require an additional 150 pieces. Since three new pieces are created for each additional line drawn, we need to add = 50 new lines after the first, or 51 lines in total. Answer: (D) 3. Since we are looking for the percentage of the whole length, we may take the side length of the square to be any convenient value, as the actual length will not affect the final answer. Let us assume that the side length of the square is. Then the diameter of the circle is also because the width of the square and the diameter of the circle are equal. Using the Pythagorean Theorem, XY = + = = 8 or XY = 8. The portion of line segment XY lying outside the circle has length XY minus the diameter of the circle, or 8. The percentage of line segment XY lying outside the circle is % 9.3%. Answer: (A) 4. 1 Breenah travels along each of the sides in a direction that is either up, down, right or left. The up sides occur every fourth segment, thus they have lengths 1, 5, 9, 13, 17, 1,... or lengths that are one more than a multiple of 4. As we see, the segment of length 1 is an up side P

112 009 Gauss Contest s Page 13 We see that the upper endpoint of each up segment is units to the left and units above the upper endpoint of the previous up segment. Thus, when Breenah is standing at the upper endpoint of the up segment of length 1, she is = 10 units to the left and = 11 units above P. The following paragraphs prove this in a more formal way. We now determine the horizontal distance from point P to the up side of length 1. Following the spiral outward from P, the first horizontal line segment moves right, or +, where the positive sign indicates movement to the right. The second horizontal segment moves left 4, or 4, where the negative sign indicates movement to the left. After these two horizontal movements, we are at the line segment of length 5, an up side. To get there, we moved a horizontal distance of (+) + ( 4) = or units to the left. The horizontal distance from P to the next up side (length 9), can be found similarly. Beginning on the segment of length 5, we are already at or units left of P and we move right 6 (or +6), then left 8 (or 8). Thus, to reach the up side with length 9, we have moved horizontally ( ) + (+6) + ( 8) = 4 or 4 units left of P. This pattern of determining the horizontal distances from P to each of the up sides is continued in the table below. We now determine the vertical distance from point P to the upper endpoint of the up side with length 1. From P, the first vertical segment moves up 1 or +1, the second moves down 3 or 3 and the third moves up 5 or +5. Therefore the vertical position of the endpoint of the up side with length 5 is (+1) + ( 3) + (+5) = +3 or 3 units above P. Similarly, we can calculate the vertical position of each of the up side endpoints relative to P and have summarized this in the table below. Side Length Horizontal Distance Vertical Distance 5 (+) + ( 4) = (+1) + ( 3) + (+5) = +3 9 ( ) + (+6) + ( 8) = 4 (+3) + ( 7) + (+9) = ( 4) + (+10) + ( 1) = 6 (+5) + ( 11) + (+13) = ( 6) + (+14) + ( 16) = 8 (+7) + ( 15) + (+17) = +9 1 ( 8) + (+18) + ( 0) = 10 (+9) + ( 19) + (+1) = +11 We now calculate the distance, d, from P to the upper endpoint, F, of the up side of length 1. F 11 d 10 Since the calculated distances are horizontal and vertical, we have created a right angle and may find the required distance using the Pythagorean Theorem. Then d = = = 1 or d = P

113 009 Gauss Contest s Page 14 If we place the spiral on an xy-plane with point P at the origin, the coordinates of the key points reveal a pattern. Side Length Endpoint Coordinates 1 (0, 1) (, 1) 3 (, ) 4 (, ) 5 (, 3) 6 (4, 3) 7 (4, 4) 8 ( 4, 4). From the table we can see that after finishing a side having length that is a multiple of 4, say 4k, we are at the point ( k, k) (the basis for this argument is shown in solution 1). Therefore, after completing the side of length 0, we are at the point ( 10, 10). All sides of length 4k travel toward the left. We must now move vertically upward 1 units from this point ( 10, 10). Moving upward, this last side of length 1 will end at the point F ( 10, 11). This point is left 10 units and up 11 units from P (0, 0). Using the Pythagorean Theorem, P F = = = 1 or P F = Answer: (B) 5. Consider all possible sums of pairs of numbers that include p: p + q, p + r, p + s, p + t and p + u. We see that p is included in 5 different sums (once with each of the other 5 numbers in the list). Similarly, each of the numbers will be included in 5 sums. If the sums of all 15 pairs are added together, each of p, q, r, s, t, and u will have been included 5 times. Therefore, 5p + 5q + 5r + 5s + 5t + 5u = = 980 or 5(p + q + r + s + t + u) = 980. Dividing this equation by 5, we obtain p + q + r + s + t + u = 980 = Since p and q are the smallest two integers, their sum must be the smallest of all of the pairs; thus, p + q = 5. Similarly, t and u are the largest two integers and their sum must be the largest of all of the pairs; thus, t + u = 117. Now, (p + q) + r + s + (t + u) = 196 becomes 5 + r + s = 196 or r + s = = 54. Answer: (B)

114 Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 008 Gauss Contests (Grades 7 and 8) Wednesday, May 14, 008 s c 007 Waterloo Mathematics Foundation

115 008 Gauss Contest s Page 3 Grade 7 1. Calculating, 6 3 = 1 3 = 9.. Calculating, = = = Answer: (A) Answer: (E) 3. Using a common denominator of 8, we have = = 7 8. Answer: (D) 4. Since the polygon has perimeter 108 cm and each side has length 1 cm, then the polygon has = 9 sides. Answer: (D) 5. In the set, three of the numbers are greater than or equal to 3, and two of the numbers are less than 3. The smallest number must be one of the numbers that is less than 3, that is,.3 or.3. Of these two numbers,.3 is the smallest, so is the smallest number in the set. Answer: (D) 6. Since P Q is a straight line, then x + x + x + x + x = 180 or 5x = 180 or x = 36. Answer: (A) 7. 0 is not a prime number, since it is divisible by. 1 is not a prime number, since it is divisible by 3. 5 is not a prime number, since it is divisible by 5. 7 is not a prime number, since it is divisible by 3. 3 is a prime number, since its only positive divisors are 1 and Kayla walked 8 km on Monday. On Tuesday, she walked 8 = 4 km. On Wednesday, she walked 4 = km. On Thursday, she walked = 1 km. On Friday, she walked 1 = 0.5 km. Answer: (C) Answer: (E) 9. Since 50% selected chocolate and 10% selected strawberry as their favourite flavour, then overall 50% + 10% = 60% chose chocolate or strawberry as their favourite flavour. Now 60% = 60 = 3, so 3 of the people surveyed selected chocolate or strawberry as their favourite flavour. Answer: (A) 10. Since Max sold 41 glasses of lemonade on Saturday and 53 on Sunday, he sold = 94 glasses in total. Since he charged 5 cents for each glass, then his total sales were 94 $0.5 = $3.50. Answer: (A)

116 008 Gauss Contest s Page Since Chris spent $68 in total and $5 on the helmet, then he spent $68 $5 = $43 on the two hockey sticks. Since the two sticks each cost the same amount, then this cost was $43 = $1.50. Answer: (C) 1. The number below and between 17 and 6 is 17 6 = 11. The number below and between 8 and 11 is 11 8 = 3. The number below and between 11 and is 11 = 9. The number below and between 7 and 3 is 7 3 = 4. The number below and between 3 and 9 is 9 3 = x Therefore, x = 6 4 =. Answer: (B) 13. Since P Q = P R, then P QR = P RQ. Since the angles in a triangle add up to 180, then 40 + P QR + P RQ = 180, so P QR + P RQ = 140. Since P QR = P RQ, then P QR = P RQ = 70. Since the angle labelled as x is opposite P RQ, then x = P RQ = 70, so x = 70. Answer: (B) 14. The sum of Wesley s and Breenah s ages is. After each year, each of their ages increases by 1, so the sum of their ages increases by. For the sum to increase from to = 44, the sum must increase by, which will take = 11 years. Answer: (E) 15. The first transformation is a 180 rotation of the letter, which gives G. The second transformation is a reflection across a vertical axis, which gives G. Answer: (D) G 16. In the diagram, the length of one side of the large square is equal to eight side lengths of the smaller squares, so the large square consists of 8 8 = 64 small squares. Of these 64 small squares, 48 are shaded. (We can obtain this number by counting the 48 shaded squares or by counting the 16 unshaded squares.) As a percentage, this fraction equals % = % = 75%. Answer: (D) G G Since the perimeter of a rectangle equals twice the length plus twice the width, then the length plus the width equals 10 = 60. Since the length equals twice the width plus 6, then twice the width plus the width equals 60 6 = 54. In other words, three times the width equals 54, so the width equals 54 3 = 18.

117 008 Gauss Contest s Page 5 Let the width of the rectangle be w. Then the length of the rectangle is w + 6. Since the perimeter is 10, then w + (w + 6) = 10 w + 4w + 1 = 10 6w = 108 w = 18 so the width is 18. Answer: (B) 18. The sum of Rishi s marks so far is = 315. Since Rishi s mark on his next test is between 0 and 100, the sum of his marks will be between = 315 and = 415 after his next test. Since his average equals the sum of his marks divided by the number of marks, then his average will be between = 63 and 5 5 = 83. Of the given choices, the only one in this range is 8. Answer: (C) 19. After some experimentation, the only way in which the two given pieces can be put together to stay within a 4 4 grid and so that one of the given choices can fit together with them is to rotate the second piece by 90 clockwise, and combine to obtain. Therefore, the missing piece is. Answer: (C) 0. The possible ways of writing 7 as the product of three different positive integers are: 1 36; 1 3 4; ; 1 6 1; 1 8 9; 3 1; 4 9; (We can find all of these possibilities systematically by starting with the smallest possible first number and working through the possible second numbers, then go to the next possible smallest first number and continue.) The sums of these sets of three numbers are 39, 8, 3, 19, 18, 17, 15, 13, so the smallest possible sum is 13. Answer: (A) 1. Since Andrea has completed 3 of the total 168 km, then she has completed km or = 7 km. This means that she has = 96 km remaining. To complete the 96 km in her 3 remaining days, she must average 96 = 3 km per day. 3 Answer: (D). 1 Since P Q is parallel to SR, then the height of P QS (considering P Q as the base) and the height of SRQ (considering SR as the base) are the same (that is, the vertical distance between P Q and SR).

118 008 Gauss Contest s Page 6 Since SR is twice the length of P Q and the heights are the same, then the area of SRQ is twice the area of P QS. In other words, the area of P QS is 1 of the total area of the trapezoid, or 1 1 = Draw a line from Q to T, the midpoint of SR. P Q S T R Since SR = (P Q) and T is the midpoint of SR, then P Q = ST = T R. We consider P Q, ST and T R as the bases of P QS, ST Q and T RQ, respectively. Using these three segments as the bases, each of these triangles has the same height, since P Q is parallel to SR. Since P Q = ST = T R and these triangles have the same height, then the three triangles each have the same area. The trapezoid is thus cut into three triangles of equal area. Therefore, the area of P QS is one-third of the area of entire trapezoid, or 1 1 = 4. 3 Answer: (B) 3. Since Ethan does not sit next to Dianne, the four must arrange themselves in one of the configurations: D E D E D E E D E D E D For each of these six configurations, there are two ways for Beverly and Jamaal to sit (either with Beverly on the left or with Jamaal on the left). Therefore, there are 6 = 1 possible ways that the four can sit. (Try listing them out!) Answer: (B) 4. Since the two large triangles are equilateral, then each of their three angles equals 60. Therefore, each of 6 small triangles in the star has an angle of 60 between the two equal sides. But each of these 6 small triangles is isosceles so each of the remaining two angles must equal 1 ( ) or 60. Therefore, each of the small triangles is equilateral. This shows us that the inner hexagon has all sides equal, and also that each angle is or 10, so the hexagon is regular. Next, we draw the three diagonals of the hexagon that pass through its centre (this is possible because of the symmetry of the hexagon).

119 008 Gauss Contest s Page 7 Also, because of symmetry, each of the angles of the hexagon is split in half, to get 10 = 60. Therefore, each of the 6 new small triangles has two 60 angles, and so must have its third angle equal to 60 as well. Thus, each of the 6 new small triangles is equilateral. So all 1 small triangles are equilateral. Since each has one side length marked by a single slash, then these 1 small triangles are all identical. Since the total area of the star is 36, then the area of each small triangle is 36 1 = 3. Since the shaded area is made up of 9 of these small triangles, its area is 9 3 = 7. Answer: (C) 5. First we look at the integers from 000 to 008. Since we can ignore the 0s when adding up the digits, the sum of all of the digits of these integers is + ( + 1) + ( + ) + ( + 3) + ( + 4) + ( + 5) + ( + 6) + ( + 7) + ( + 8) = 54 Next, we look at the integers from 1 to Again, since we can ignore digits of 0, we consider these numbers as 0001 to 1999, and in fact as the integers from 0000 to 1999, including 0000 to make 000 integers in total. Of these 000 integers, 00 have a units digit of 0, 00 have a units digit of 1, and so on. (One integer out of every 10 has a units digit of 0, and so on.) Therefore, the sum of the units digits of these integers is 00(0)+00(1)+ +00(8)+00(9) = = 9000 Of these 000 integers, 00 have a tens digit of 0, 00 have a tens digit of 1, and so on. (Ten integers out of every 100 have a tens digit of 0, and so on.) Therefore, the sum of the tens digits of these integers is 00(0) + 00(1) (8) + 00(9) = 9000 Of these 000 integers, 00 have a hundreds digit of 0 (that is, 0000 to 0099 and 1000 to 1099), 00 have a hundreds digit of 1, and so on. (One hundred integers out of every 1000 have a hundreds digit of 0, and so on.) Therefore, the sum of the hundreds digits of these integers is 00(0) + 00(1) (8) + 00(9) = 9000 Of these 000 integers, 1000 have a thousands digit of 0 and 1000 have a thousands digits of 1. Therefore, the sum of the thousands digits of these integers is 1000(0) (1) = 1000 Overall, the sum of all of the digits of these integers is = Answer: (E)

120 008 Gauss Contest s Page 8

121 008 Gauss Contest s Page 9 Grade 8 1. Using the correct order of operations, 8 (6 4) + = 8 + = 16 + = 18. Answer: (C). Since the polygon has perimeter 108 cm and each side has length 1 cm, then the polygon has = 9 sides. Answer: (D) 3. Since P QR = 90, then x + x = 90 or 3x = 90 or x = Calculating, (1 + ) (1 + ) = 3 (1 + 4) = 9 5 = 4. Answer: (A) Answer: (B) 5. When these four numbers are listed in increasing order, the two negative numbers come first, followed by the two positive numbers. Of the two positive numbers, 0.8 and.8, the number 0.8 is the smallest. Of the two negative numbers, 0. and 8., the number 8. is the smallest. Therefore, the correct order is 8., 0., 0.8,.8. Answer: (A) 6. From the given formula, the number that should be placed in the box is = 15+4 = 19. Answer: (E) 7. Since 50% selected chocolate and 10% selected strawberry as their favourite flavour, then overall 50% + 10% = 60% chose chocolate or strawberry as their favourite flavour. Now 60% = 60 = 3, so 3 of the people surveyed selected chocolate or strawberry as their favourite flavour. Answer: (A) 8. 1 Since 5 times the number minus 9 equals 51, then 5 times the number must equal 60 (that is, ). Therefore, the original number is 60 divided by 5, or 1. Let the original number be x. Then 5x 9 = 51, so 5x = = 60, so x = 60 5 = 1. Answer: (D) 9. 1 Since Danny weighs 40 kg, then 0% of his weight is 0 40 = 1 40 = 8 kg Since Steven weighs 0% more than Danny, his weight is = 48 kg. Since Steven weighs 0% more than Danny, then Steven s weight is 10% of Danny s weight. Since Danny s weight is 40 kg, then Steven s weight is = 6 40 = 48 kg Answer: (C)

122 008 Gauss Contest s Page Of the given 11 numbers, the numbers 3, 5, 7, 11 and 13 are prime. (4, 6, 8, 10 and 1 are not prime, since they are divisible by, and 9 is not prime since it is divisible by 3.) Therefore, 5 of the 11 numbers are prime. Thus, if a card is chosen at random and flipped over, the probability that the number on this card is a prime number is Answer: (E) 11. In centimetres, the dimensions of the box are 0 cm, 50 cm, and 100 cm (since 1 m equals 100 cm). Therefore, the volume of the box is (0 cm) (50 cm) (100 cm) = cm 3 Answer: (D) 1. 1 Since each pizza consists of 8 slices and each slice is sold for $1, then each pizza is sold for $8 in total. Since 55 pizzas are sold, the total revenue is 55 $8 = $440. Since 55 pizzas were bought initially, the total cost was 55 $6.85 = $ Therefore, the total profit was $440 $ = $63.5. Since each pizza consists of 8 slices and each slice is sold for $1, then each pizza is sold for $8 in total. Since each pizza was bought for $6.85 initially, then the school makes a profit of $8.00 $6.85 or $1.15 per pizza. Since the school completely sold 55 pizzas, then its total profit was 55 $1.15 = $63.5. Answer: (D) 13. Since RSP is a straight line, then RSQ + QSP = 180, so RSQ = = 100. Since RSQ is isosceles with RS = SQ, then RQS = 1 (180 RSQ) = 1 ( ) = 40 Similarly, since P SQ is isosceles with P S = SQ, then P QS = 1 (180 P SQ) = 1 ( ) = 50 Therefore, P QR = P QS + RQS = = 90. Answer: (B) 14. On Monday, Amos read 40 pages. On Tuesday, Amos read 60 pages, for a total of = 100 pages so far. On Wednesday, Amos read 80 pages, for a total of = 180 pages so far. On Thursday, Amos read 100 pages, for a total of = 80 pages so far. On Friday, Amos read 10 pages, for a total of = 400 pages so far. Therefore, Amos finishes the 400 page book on Friday. Answer: (A)

123 008 Gauss Contest s Page If Abby had 3 nickels, the total value would be 3 $0.05 = $1.15. But the total value of Abby s coins is $4.55, which is $3.40 more. Since a quarter is worth 0 cents more than a nickel, then every time a nickel is replaced by a quarter, the total value of the coins increase by 0 cents. For the total value to increase by $3.40, we must replace $3.40 $0.0 = 17 nickels with quarters. Therefore, Abby has 17 quarters. (To check, if Abby has 17 quarters and 6 nickels, the total value of the coins that she has is 17 $ $0.05 = $4.5 + $0.30 = $4.55.) Answer: (B) 16. After some experimentation, the only way in which the two given pieces can be put together to stay within a 4 4 grid and so that one of the given choices can fit together with them is to rotate the second piece by 90 clockwise, and combine to obtain. Therefore, the missing piece is. Answer: (C) 17. The digits after the decimal point occur in repeating blocks of 6 digits. Since = , then the 008th digit after the decimal point occurs after 334 blocks of digits have been used. In 334 blocks of 6 digits, there are = 004 digits in total. Therefore, the 008th digit is 4 digits into the 335th block, so must be 8. Answer: (A) 18. Since Andrea has completed 3 of the total 168 km, then she has completed km or = 7 km. This means that she has = 96 km remaining. To complete the 96 km in her 3 remaining days, she must average 96 = 3 km per day. 3 Answer: (D) After some trial and error, you might discover that x = 0 and y = 7 works, since = Therefore, since we are asked for the unique value of y x, it must be 7 0 = 7. When performing this addition, in the units column either y + 3 = x or y + 3 = x with a carry of 1, meaning that y + 3 = 10 + x. Therefore, either y x = 3 or y x = 10 3 = 7. If there was no carry, then adding up the tens digits, we would get x + x ending in a 1, which is impossible as x + x = x which is even. Therefore, the addition y + 3 must have a carry of 1. Therefore, y x = 7. Answer: (C)

124 008 Gauss Contest s Page The area of a trapezoid equals one-half times the sum of the bases times the height. Therefore, the area of this trapezoid is We draw diagonal BD. 1 (9 + 11) 3 = = 10 3 = 30 A 3 9 B 5 D 11 C ABD has a base of 9 and a height of 3. BCD has a base of 11 and a height of 3. The area of the trapezoid is equal to the sum of the areas of these two triangles, or = = 60 = 30 3 Draw perpendicular lines from B to CD and from D to AB, as shown. A Q 3 B 5 D P C By the Pythagorean Theorem, BP + P C = BC or 3 + P C = 5 or P C = 5 9 = 16, so P C = 4. Since DC = 11, then DP = 11 4 = 7. Since QBP D is a rectangle, then QB = DP = 7, so AQ = 9 7 =. The area of trapezoid ABCD equals the sum of the area of AQD, rectangle QBP D and BP C, or = = 30 Answer: (D) 1. The object has 7 front faces, each of which is 1 1. Therefore, the surface area of the front is = 7. Similarly, the surface area of the back is 7. Now consider the faces on the left, top, right and bottom. Each of these faces is 1, so each has an area of. How many of these faces are there? If we start at the bottom left and travel clockwise around the figure, we have left faces, top faces, left faces, 1 top face, 3 right faces, 1 bottom face, 1 right face, and bottom faces, or 14 faces in total. Therefore, the surface area accounted for by these faces is 14 = 8. Therefore, the total surface area of the object is = 4. Answer: (A)

125 008 Gauss Contest s Page 13. There are 6 possibilities for the first row of the grid: 1,, 3 1, 3,, 1, 3, 3, 1 3, 1, 3,, 1 Consider the first row of 1,, 3: 1 3. The first column could be 1,, 3 or 1, 3, : or Each of these grids can be finished with the given rules, but can only be finished in one way. (In the first grid, the middle number in the bottom row cannot be or 3, so is 1, so the middle number in the middle row is 3, so the right column is 3, 1,. Similarly, in the second grid, the middle number in the middle row must be 1. Try completing this grid!) Therefore, a first row of 1,, 3 gives two possible grids. Similarly, each of the other 5 possible first rows will give two other grids. (We can see this by trying each of these possibilities or by for example switching all of the s and 3s to get the grids with a first row of 1, 3,.) Therefore, the total number of different ways of filling the grid is 6 = 1. Answer: (B) 3. Since the area of the larger circle is 64π and each circle is divided into two equal areas, then the larger shaded area is 1 of 64π, or 3π. Let r be the radius of the larger circle. Since the area of the larger circle is 64π, then πr = 64π or r = 64 or r = 64 = 8, since r > 0. Since the smaller circle passes through the centre of the larger circle and just touches the outer circle, then by symmetry, its diameter must equal the radius of the larger circle. (In other words, if we join the centre of the larger circle to the point where the two circles just touch, this line will be a radius of the larger circle and a diameter of the smaller circle.) Therefore, the diameter of the smaller circle is 8, so its radius is 4. Therefore, the area of the smaller circle is π(4 ) = 16π, so the smaller shaded area is 1 16π or 8π. Therefore, the total of the shaded areas is 3π + 8π = 40π. Answer: (D) 4. First we look at the integers from 000 to 008. Since we can ignore the 0s when adding up the digits, the sum of all of the digits of these integers is + ( + 1) + ( + ) + ( + 3) + ( + 4) + ( + 5) + ( + 6) + ( + 7) + ( + 8) = 54 Next, we look at the integers from 1 to Again, since we can ignore digits of 0, we consider these numbers as 0001 to 1999, and in fact as the integers from 0000 to Of these 000 integers, 00 have a units digit of 0, 00 have a units digit of 1, and so on. (One integer out of every 10 has a units digit of 0, and so on.) Therefore, the sum of the units digits of these integers is 00(0) + 00(1) (8) + 00(9) = 00( ) = 00(45) = 9000.

126 008 Gauss Contest s Page 14 Of these 000 integers, 00 have a tens digit of 0, 00 have a tens digit of 1, and so on. (Ten integers out of every 100 have a tens digit of 0, and so on.) Therefore, the sum of the tens digits of these integers is 00(0) + 00(1) (8) + 00(9) = 9000 Of these 000 integers, 00 have a hundreds digit of 0 (that is, 0000 to 0099 and 1000 to 1099), 00 have a hundreds digit of 1, and so on. (One hundred integers out of every 1000 have a hundreds digit of 0, and so on.) Therefore, the sum of the hundreds digits of these integers is 00(0) + 00(1) (8) + 00(9) = 9000 Of these 000 integers, 1000 have a thousands digit of 0 and 1000 have a thousands digits of 1. Therefore, the sum of the thousands digits of these integers is 1000(0) (1) = 1000 Overall, the sum of all of the digits of these integers is = Answer: (E) 5. Since the length of the candles was equal at 9 p.m., the longer one burned out at 10 p.m., and the shorter one burned out at midnight, then it took 1 hour for the longer candle and 3 hours for the shorter candle to burn this equal length. Therefore, the longer candle burned 3 times as quickly as the shorter candle. Suppose that the shorter candle burned x cm per hour. Then the longer candle burned 3x cm per hour. From its lighting at 3 p.m. to 9 p.m., the longer candles burned for 6 hours, so burned 6 3x or 18x cm. From its lighting at 7 p.m. to 9 p.m., the shorter candle burned for hours, so burns x = x cm. But, up to 9 p.m., the longer candle burned 3 cm more than the shorter candle, since it began 3 cm longer. Therefore, 18x x = 3 or 16x = 3 or x =. In summary, the shorter candle burned for 5 hours at cm per hour, so its initial length was 10 cm. Also, the longer candle burned for 7 hours at 6 cm per hour, so its initial length was 4 cm. Thus, the sum of the original lengths is = 5 cm. Answer: (E)

127 Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 007 Gauss Contests (Grades 7 and 8) Wednesday, May 16, 007 s c 006 Waterloo Mathematics Foundation

128 007 Gauss Contest s Page 3 Grade 7 1. Calculating, (4 3) = 1 =.. Since one thousand is 1000, then ten thousand is Answer: (B) Answer: (C) 3. When we subtract 5 from the missing number, the answer is, so to find the missing number, we add 5 to and obtain 7. (Check: 7 5 =.) Answer: (A) 4. 1 As a fraction, 80% is 80 or Therefore, Mukesh got 4 of the possible 50 marks, or 4 50 = 40 marks. 5 5 Since Mukesh got 80% of the 50 marks, he got = 80 = 40 marks. Answer: (A) = = = (using a common denominator) = (converting each fraction to a decimal) = Answer: (D) 6. 1 Mark has 3 of a dollar, or 75 cents. 4 Carolyn has 3 of a dollar, or 30 cents. 10 Together, they have = 105 cents, or $1.05. Since Mark has 3 of a dollar and Carolyn has 3 of a dollar, then together they have = = 1 of a dollar Since is equivalent to, they have $ Answer: (E) 7. From the graph, the student who ate the most apples ate 6 apples, so Lorenzo ate 6 apples. Also from the graph, the student who ate the fewest apples ate 1 apple, so Jo ate 1 apple. Therefore, Lorenzo ate 6 1 = 5 more apples than Jo. Answer: (B)

129 007 Gauss Contest s Page 4 8. Since the angles in a triangle add to 180, then the missing angle in the triangle is = 70. We then have: C 50 A X 70 x 60 B D Since BXC = 70, then AXC = 180 BXC = 110. Since AXC = 110, then DXA = 180 AXC = 70. Therefore, x = 70. (Alternatively, we could note that when two lines intersect, the vertically opposite angles are equal so DXA = BXC = 70.) Answer: (E) 9. When the word BANK is viewed from the inside of the window, the letters appear in the reverse order and the letters themselves are all backwards, so the word appears as. Answer: (D) 10. Since a large box costs $3 more than a small box and a large box and a small box together cost $15, then replacing the large box with a small box would save $3. This tells us that two small boxes together cost $1. Therefore, one small box costs $6. Answer: (D) 11. Since each number in the Fibonacci sequence, beginning with the, is the sum of the two previous numbers, then the sequence continues as 1, 1,, 3, 5, 8, 13, 1. Thus, 1 appears in the sequence. Answer: (B) 1. The probability that Mary wins the lottery is equal to the number of tickets that Mary bought divided by the total number of tickets in the lottery. We are told that the probability that Mary wins is Since there were 10 tickets in total sold, we would like to write 1 as a fraction with 10 in 15 the denominator. 1 Since = 8, then we need to multiply the numerator and denominator of each by 8 15 to obtain a denominator of 10. Therefore, the probability that Mary wins is = 8. Since there were 10 tickets sold, 10 then Mary must have bought 8 tickets. Answer: (D) BANK

130 007 Gauss Contest s Page We look at each of the choices and try to make them using only 3 cent and 5 cent stamps: (A): 7 cannot be made, since no more than one 5 cent and two 3 cent stamps could be used (try playing with the possibilities!) (B): 13 = (C): 4 cannot be the answer since a larger number (7) already cannot be made (D): 8 = (E): 9 = Therefore, the answer must be 7. (We have not really justified that 7 is the largest number that cannot be made using only 3s and 5s; we have, though, determined that 7 must be the answer to this question, since it is the only possible answer from the given possibilities! See for a justification that 7 is indeed the answer.) We make a table to determine which small positive integers can be made using 3s and 5s: Integer Combination 1 Cannot be made Cannot be made Cannot be made Cannot be made Every integer larger than 11 can also be made because the last three integers in our table can be made and we can add a 3 to our combinations for 9, 10 and 11 to get combinations for 1, 13 and 14, and so on. From the table, the largest amount of postage that cannot be made is 7. Answer: (A) 14. We list all of the possible orders of finish, using H, R and N to stand for Harry, Ron and Neville. The possible orders are HNR, HRN, NHR, NRH, RHN, RNH. (It is easiest to list the orders in alphabetical order to better keep track of them.) There are 6 possible orders. Answer: (B) The positive whole numbers that divide exactly into 40 are 1,, 4, 5, 8, 10, 0, 40. The positive whole numbers that divide exactly into 7 are 1,, 3, 4, 6, 8, 9, 1, 18, 4, 36, 7. The numbers that occur in both lists are 1,, 4, 8, or four numbers in total.

131 007 Gauss Contest s Page 6 The greatest common divisor of 40 and 7 is 8. Any common divisor of 40 and 7 is a divisor of the greatest common divisor (namely 8) and vice-versa. Since the positive divisors of 8 are 1,, 4, and 8, there are four such common positive divisors. Answer: (C) 16. The first scale tells us that a square and a circle together have a mass of 8. The second scale tells us that a square and two circles together have a mass of 11. We can replace the square and one circle on the second scale with an 8, so 8 plus the mass of a circle gives a mass of 11. This tells us that the mass of a circle is 3. From the third scale, since the mass of a circle and two triangles is 15, then the mass of the two triangles only is 15 3 = 1. Therefore, the mass of one triangle is 1 = 6. Answer: (D) 17. The total cost to use the kayak for 3 hours is 3 $5 = $15. Since the total rental cost for 3 hours is $30, then the fixed fee to use the paddle is $30 $15 = $15. For a six hour rental, the total cost is thus $15 + (6 $5) = $15 + $30 = $45. Answer: (C) Julie s birthday was = 104 days before Fred s birthday. When we divide 104 by 7 (the number of days in one week), we obtain a quotient of 14 and a remainder of 6. In 14 weeks, there are 14 7 = 98 days, so 98 days before Fred s birthday was also a Monday. Since Julie s birthday was 104 days before Fred s, this was 6 days still before the Monday 98 days before Fred s birthday. The 6th day before a Monday is a Tuesday. Therefore, Julie s birthday was a Tuesday. 37 days is 5 weeks plus days. Since Fred s birthday was on a Monday and Pat s birthday was 37 days before Fred s, then Pat s birthday was on a Saturday. 67 days is 9 weeks plus 4 days. Since Pat s birthday was on a Saturday and Julie s birthday was 67 days before Pat s, then Julie s birthday was on a Tuesday. Answer: (D) 19. The positive whole numbers less than 1000 that end with 77 are 77, 177, 77, 377, 477, 577, 677, 777, 877, 977. The positive whole numbers less than 1000 which begin with 77 are 77, 770, 771, 77, 773, 774, 775, 776, 777, 778, 779. There is no other way for a positive whole number less than 1000 to contain at least two 7 s side-by-side. There are 10 numbers in the first list and 11 numbers in the second list. Since numbers appear in both lists, the total number of whole numbers in the two lists is = 19. Answer: (E)

132 007 Gauss Contest s Page 7 0. Since the perimeter of the square is 48, its side length is 48 4 = 1. Since the side length of the square is 1, its area is 1 1 = 144. The area of the triangle is 1 48 x = 4x. Since the area of the triangle equals the area of the square, then 4x = 144 or x = 6. Answer: (C) 1. 1 Starting at the K there are two possible paths that can be taken. At each A, there are again two possible paths that can be taken. Similarly, at each R there are two possible paths that can be taken. Therefore, the total number of paths is = 8. (We can check this by actually tracing out the paths.) Each path from the K at the top to one of the L s at the bottom has to spell KARL. There is 1 path that ends at the first L from the left. This path passes through the first A and the first R. There are 3 paths that end at the second L. The first of these passes through the first A and the first R. The second of these passes through the first A and the second R. The third of these passes through the second A and the second R. There are 3 paths that end at the third L. The first of these passes through the first A and the second R. The second of these passes through the second A and the second R. The third of these passes through the second A and the third R. There is 1 path that ends at the last L. This path passes through the last A and the last R. So the total number of paths to get to the bottom row is = 8, which is the number of paths that can spell KARL. Answer: (D). Since the average of four numbers is 4, their sum is 4 4 = 16. For the difference between the largest and smallest of these numbers to be as large as possible, we would like one of the numbers to be as small as possible (so equal to 1) and the other (call it B for big) to be as large as possible. Since one of the numbers is 1, the sum of the other three numbers is 16 1 = 15. For the B to be as large as possible, we must make the remaining two numbers (which must be different and not equal to 1) as small as possible. So these other two numbers must be equal to and 3, which would make B equal to 15 3 = 10. So the average of these other two numbers is + 3 = 5 or 1. Answer: (B) 3. 1 Since we are dealing with fractions of the whole area, we may make the side of the square any convenient value. Let us assume that the side length of the square is 4. Therefore, the area of the whole square is 4 4 = 16. The two diagonals of the square divide it into four pieces of equal area (so each piece has area 16 4 = 4). The shaded area is made up from the right quarter of the square with a small triangle removed, and so has area equal to 4 minus the area of this small triangle.

133 007 Gauss Contest s Page 8 This small triangle is half of a larger triangle. This larger triangle has its base and height each equal to half of the side length of the square (so equal to ) and has a right angle. So the area of this larger triangle is 1 =. So the area of the small triangle is 1 = 1, and so the area of the shaded region is 4 1 = 3. Therefore, the shaded area is 3 of the area of the whole square. 16 Draw a horizontal line from the centre of the square through the shaded region. The two diagonals divide the square into four pieces of equal area. The new horizontal line divides one of these pieces into two parts of equal area. Therefore, the shaded region above the new horizontal line is 1 1 = 1 of the total area of the square. 4 8 The shaded piece below this new horizontal line is half of the bottom right part of this right-hand piece of the square. (It is half of this part because the shaded triangle and unshaded triangle making up this part have the same shape.) So this remaining shaded piece is 1 1 = 1 of the 8 16 total area of the square. In total, the shaded region is = 3 of the total area of the square Answer: (C) 4. First, we try to figure out what digit Q is. Since the product is not equal to 0, Q cannot be 0. Since the product has four digits and the top number has three digits, then Q (which is multiplying the top number) must be bigger than 1. Looking at the units digits in the product, we see that Q Q has a units digit of Q. Since Q > 1, then Q must equal 5 or 6 (no other digit gives itself as a units digit when multiplied by itself). But Q cannot be equal to 5, since if it was, the product RQ5Q would end 55 and each of the two parts (P P Q and Q) of the product would end with a 5. This would mean that each of the parts of the product was divisible by 5, so the product should be divisible by 5 5 = 5. But a number ending in 55 is not divisible by 5. Therefore, Q = 6. So the product now looks like P P 6 6 R Now when we start the long multiplication, 6 6 gives 36, so we write down 6 and carry a 3. When we multiply P 6 and add the carry of 3, we get a units digit of 5, so the units digit of

134 007 Gauss Contest s Page 9 P 6 should be. For this to be the case, P = or P = 7. We can now try these possibilities: 6 6 = 1356 and = Only the second ends 656 like the product should. So P = 7 and R = 4, and so P + Q + R = = 17. Answer: (E) 5. The easiest way to keep track of the letters here is to make a table of what letters arrive at each time, what letters are removed, and what letters stay in the pile. Time Letters Letters Remaining Pile Arrived Removed (bottom to top) 1:00 1,, 3 3, 1 1:05 4, 5, 6 6, 5 1, 4 1:10 7, 8, 9 9, 8 1, 4, 7 1:15 10, 11, 1 1, 11 1, 4, 7, 10 1:0 13, 14, 15 15, 14 1, 4, 7, 10, 13 1:5 16, 17, 18 18, 17 1, 4, 7, 10, 13, 16 1:30 19, 0, 1 1, 0 1, 4, 7, 10, 13, 16, 19 1:35, 3, 4 4, 3 1, 4, 7, 10, 13, 16, 19, 1:40 5, 6, 7 7, 6 1, 4, 7, 10, 13, 16, 19,, 5 1:45 8, 9, 30 30, 9 1, 4, 7, 10, 13, 16, 19,, 5, 8 1:50 31, 3, 33 33, 3 1, 4, 7, 10, 13, 16, 19,, 5, 8, 31 1:55 34, 35, 36 36, 35 1, 4, 7, 10, 13, 16, 19,, 5, 8, 31, 34 1:00 None 34, 31 1, 4, 7, 10, 13, 16, 19,, 5, 8 1:05 None 8, 5 1, 4, 7, 10, 13, 16, 19, 1:10 None, 19 1, 4, 7, 10, 13, 16 1:15 None 16, 13 1, 4, 7, 10 (At 1:55, all 36 letters have been delivered, so starting at 1:00 letters are only removed and no longer added.) Letter #13 is removed at 1:15. Answer: (A)

135 007 Gauss Contest s Page 10 Grade 8 1. Calculating, (4 1) (4 + 1) = = 3. Answer: (E). Converting to decimals, = = Answer: (B) 3. We use the graph to determine the difference between the high and low temperature each day. Day High Low Difference Monday Tuesday Wednesday Thursday Friday Therefore, the difference was the greatest on Thursday. Answer: (D) 4. When the cube is tossed, the total number of possibilities is 6 and the number of desired outcomes is. So the probability of tossing a 5 or 6 is 6 or 1 3. Answer: (C) 5. Since the side length of the cube is x cm, its volume is x 3 cm 3. Since the volume is known to be 8 cm 3, then x 3 = 8 so x = (the cube root of 8). Answer: (A) 6. Since a 3 minute phone call costs $0.18, then the rate is $ = $0.06 per minute. For a 10 minute call, the cost would be 10 $0.06 = $0.60. Answer: (B) 7. Since there are 1000 metres in a kilometre, then 00 metres is equivalent to 00 km or 0. km Answer: (A) 8. The children in the Gauss family have ages 7, 7, 7, 14, The mean of their ages is thus = = 10. Answer: (E) 9. Since x = 5 and y = x + 3, then y = = 8. Since y = 8 and z = 3y + 1, then z = 3(8) + 1 = = 5. Answer: (B) 10. The possible three-digit numbers that can be formed using the digits 5, 1 and 9 are: 519, 591, 951, 915, 195, 159. The largest of these numbers is 951 and the smallest is 159. The difference between these numbers is = 79. Answer: (C)

136 007 Gauss Contest s Page Since Lily is 90 cm tall, Anika is 4 as tall as Lily, and Sadaf is is = = = 150 cm tall as tall as Anika, then Sadaf Since Lily is 90 cm tall and Anika is 4 of her height, then Anika is = = 10 cm tall Since Anika is 10 cm tall and Sadaf is 5 of her height, then Sadaf is = = 150 cm tall. Answer: (E) 1. 1 Since BCA = 40 and ADC is isosceles with AD = DC, then DAC = ACD = 40. Since the sum of the angles in a triangle is 180, then ADC = 180 DAC ACD = = 100. Since ADB and ADC are supplementary, then ADB = 180 ADC = = 80. Since ADB is isosceles with AD = DB, then BAD = ABD. Thus, BAD = 1 (180 ADB) = 1 ( ) = 1 (100 ) = 50. Therefore, BAC = BAD + DAC = = 90. Since ABD and ACD are isosceles, then BAD = ABD and DAC = ACD. These four angles together make up all of the angles of ABC, so their sum is 180. Since BAC is half of the the sum of these angles (as it incorporates one angle of each pair), then BAC = 1 (180 ) = 90. Answer: (D) 13. For each of the art choices, Cayli can chose 1 of 3 sports choices and 1 of 4 music choices. So for each of the art choices, Cayli has 3 4 = 1 possible combinations of sports and music. Since Cayli has art choices, her total number of choices is 1 = 4. Answer: (B) 14. At the 007 Math Olympics, Canada won 17 of 100 possible medals, or 0.17 of the possible medals. We convert each of the possible answers to a decimal and see which is closest to 0.17: (A) 1 4 = 0.5 (B) 1 5 = 0. (C) 1 6 = (D) 1 7 = (E) 1 8 = 0.15 The choice that is closest to 0.17 is 1, or (C). 6 Answer: (C) 15. Let us try the integers 5, 6, 7, 8. When 5 is divided by 4, the quotient is 1 and the remainder is 1. When 6 is divided by 4, the quotient is 1 and the remainder is. When 7 is divided by 4, the quotient is 1 and the remainder is 3. When 8 is divided by 4, the quotient is and the remainder is 0. The sum of these remainders is = 6. (When any four consecutive integers are chosen, one will have a remainder 1, one a remainder of, one a remainder of 3 and one a remainder of 0 when divided by 4.) Answer: (A)

137 007 Gauss Contest s Page Suppose that the initial radius of the circle is 1. Then its initial area is π(1) = π and its initial circumference is π(1) = π. When the radius is tripled, the new radius is 3. The new area is π(3) = 9π and the new circumference is π(3) = 6π so the area is 9 times as large and the circumference is 3 times as large. Answer: (A) 17. Since each number of votes in this problem is a multiple of 1000, we consider the number of thousands of votes that each potential Idol received, to make the numbers easier with which to work. There was a total of 519 thousand votes cast. Suppose that the winner received x thousand votes. Then his opponents received x, x 30 and x 73 thousand votes. Equating the total numbers of thousand of votes, x + (x ) + (x 30) + (x 73) = 519 4x 15 = 519 4x = 5344 x = 1336 Therefore, the winner received votes. Answer: (D) 18. When the number n is doubled, n is obtained. When y is added, n + y is obtained. When this number is divided by, we obtain 1 (n + y) = n + y. When n is subtracted, y is obtained. Answer: (E) 19. To make a fraction as large as possible, we should make the numerator as large as possible and the denominator as small as possible. Of the four numbers in the diagram, z is the largest and w is the smallest, so the largest possible fraction is z w Lorri s 40 km trip to Waterloo at 10 km/h took = hours. Lorri s 40 km trip home at 80 km/h took = 3 hours. In total, Lorri drove 480 km in 5 hours, for an average speed of 480 = 96 km/h. 5 Answer: (E) Lorri s 40 km trip to Waterloo at 10 km/h took = hours. Lorri s 40 km trip home at 80 km/h took = 3 hours. Over the 5 hours that Lorri drove, her speeds were 10, 10, 80, 80, and 80, so her average speed was = = 96 km/h. Answer: (B)

138 007 Gauss Contest s Page The area of rectangle W XY Z is 10 6 = 60. Since the shaded area is half of the total area of W XY Z, its area is 1 60 = 30. Since AD and W X are perpendicular, then the shaded area has four right angles, so is a rectangle. Since square ABCD has a side length of 6, then DC = 6. Since the shaded area is 30, then P D DC = 30 or P D 6 = 30 or P D = 5. Since AD = 6 and P D = 5, then AP = 1. Answer: (A). When Chuck has the leash extended to its full length, he can move in a 70 arc, or 3 of a full 4 circle about the point where the leash is attached. (He is blocked from going further by the shed.) Shed 3 3 The area that he can play inside this circle is 3 of the area of a full circle of radius 3, or π(3 ) = 7π. 4 When the leash is extended fully to the left, Chuck just reaches the top left corner of the shed, so can go no further. When the leash is extended fully to the bottom, Chuck s leash extends 1 m below the length of the shed. This means that Chuck can play in more area to the left. 3 Shed 3 1 This area is a 90 sector of a circle of radius 1, or 1 of this circle. So this additional area is π(1 ) = 1π. 4 So the total area that Chuck has in which to play is 7π + 1π = 8π = 7π Answer: (A) 3. 1 Using $5 bills, any amount of money that is a multiple of 5 (that is, ending in a 5 or a 0) can be made. In order to get to $07 from a multiple of 5 using only $ coins, the multiple of $5 must end in a 5. (If it ended in a 0, adding $ coins would still give an amount of money that was an even integer, and so couldn t be $07.) Also, from any amount of money ending in a 5 that is less than $07, enough $ coins can always be added to get to $07.

139 007 Gauss Contest s Page 14 The positive multiples of 5 ending in a 5 that are less than 07 are 5, 15, 5,..., 195, 05. An easy way to count the numbers in this list is to remove the units digits (that is, the 5s) leaving 0, 1,,..., 19, 0; there are 1 numbers in this list. These are the only 1 multiples of 5 from which we can use $ coins to get to $07. So there are 1 different ways to make $07. We are told that 1 $ coin and 41 $5 bills make $07. We cannot use fewer $ coins, since 0 $ coins would not work, so we can only use more $ coins. To do this, we need to make change that is, trade $5 bills for $ coins. We cannot trade 1 $5 bill for $ coins, since 5 is not even. But we can trade $5 bills for 5 $ coins, since each is worth $10. Making this trade once gets 6 $ coins and 39 $5 bills. Making this trade again get 11 $ coins and 37 $5 bills. We can continue to do this trade until we have only 1 $5 bill remaining (and so $0 in $ coins, or 101 coins). So the possible numbers of $5 bills are 41, 39, 37,..., 3, 1. These are all of the odd numbers from 1 to 41. We can quickly count these to get 1 possible numbers of $5 bills and so 1 possible ways to make $07. Answer: (E) 4. To get from (, 1) to (1, 1), we go 10 units to the right and 0 units up, so we go units up every time we go 1 unit to the right. This means that every time we move 1 unit to the right, we arrive at a lattice point. So the lattice points on this segment are (, 1), (3, 3), (4, 5), (5, 7), (6, 9), (7, 11), (8, 13), (9, 15), (10, 17), (11, 19), (1, 1) (There cannot be more lattice points in between as we have covered all of the possible x- coordinates.) To get from (, 1) to (17, 6), we go 15 units to the right and 5 units up, so we go 3 units to the right every time we go 1 unit up. This means that every time we move 1 unit up, we arrive at a lattice point. So the lattice points on this segment are (, 1), (5, ), (8, 3), (11, 4), (14, 5), (17, 6) To get from (1, 1) to (17, 6), we go 5 units to the right and 15 units down, so we go 3 units down every time we go 1 unit to the right. This means that every time we move 1 unit to the right, we arrive at a lattice point. So the lattice points on this segment are (1, 1), (13, 18), (14, 15), (15, 1), (16, 9), (17, 1) In total, there are = 3 points on our three lists. But 3 points (the 3 vertices of the triangle) have each been counted twice, so there are in fact 3 3 = 0 different points on our lists. Thus, there are 0 lattice points on the perimeter of the triangle. Answer: (C)

140 007 Gauss Contest s Page To find the area of quadrilateral DRQC, we subtract the area of P RQ from the area of P DC. First, we calculate the area of P DC. We know that DC = AB = 5 cm and that DCP = 90. When the paper is first folded, P C is parallel to AB and lies across the entire width of the paper, so P C = AB = 5 cm. Therefore, the area of P DC is = 5 = 1.5 cm. Next, we calculate the area of P RQ. We know that P DC has P C = 5 cm, P CD = 90, and is isosceles with P C = CD. Thus, DP C = 45. Similarly, ABQ has AB = BQ = 5 cm and BQA = 45. Therefore, since BC = 8 cm and P B = BC P C, then P B = 3 cm. Similarly, AC = 3 cm. Since P Q = BC BP QC, then P Q = cm. Also, RP Q = DP C = 45 and RQP = BQA = 45. Q P 45 R 45 Using four of these triangles, we can create a square of side length cm (thus area 4 cm ). The area of one of these triangles (for example, P RQ) is 1 4 of the area of the square, or 1 cm. So the area of quadrilateral DRQC is therefore = 11.5 cm. Answer: (D)

141 Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 006 Gauss Contests (Grades 7 and 8) Wednesday, May 10, 006 s c 005 Waterloo Mathematics Foundation

142 006 Gauss Contest s Page 3 Grade 7 1. Calculating, (8 4) + 3 = = 35. Answer: (D). Since the angles along a straight line add to 180, then x + 40 = 180 or x + 40 = 180 or x = 140. Answer: (B) 3. To determine the number of $50 bills, we divide the total amount of money by 50, to get = 00 bills. Therefore, Mikhail has 00 $50 bills. Answer: (B) 4. The figure has 8 sides, each of equal length. Since the length of each side is, then the perimeter of the figure is 8 = 16. Answer: (A) 5. Using a common denominator, = = Answer: (C) 6. Calculating by determining each product first, = = Answer: (C) 7. Since 3 + 5x = 8, then 5x = 8 3 = 5 so x = 5 5 = 5. Answer: (C) 8. Calculating, 9 9 = = 81 3 = 78. Answer: (E) 9. In total, there are = 11 balls in the bag. Since there are 5 yellow balls, then the probability of choosing a yellow ball is 5 11 Ȧnswer: (B) 10. Since the left edge of the block is at the 3 on the ruler and the right edge of the block is between the 5 and 6, then the length of the block is between and 3. Looking at the possible choices, the only choice between and 3 is (C) or.4 cm. (Looking again at the figure, the block appears to end roughly halfway between the 5 and the 6, so.4 cm is reasonable.) Answer: (C) Since the sales tax is 15%, then the total price for the CD including tax is 1.15 $14.99 = $ which rounds to $17.4. Since the sales tax is 15%, then the amount of tax on the CD which costs $14.99 is 0.15 $14.99 = $.485, which rounds to $.5. Therefore, the total price of the CD including tax is $ $.5 = $17.4. Answer: (A)

143 006 Gauss Contest s Page Since the pool has dimensions 6 m by 1 m by 4 m, then its total volume is = 88 m 3. Since the pool is only half full of water, then the volume of water in the pool is 1 88 m3 or 144 m 3. Since the pool is half full of water, then the depth of water in the pool is 1 4 = m. Therefore, the portion of the pool which is filled with water has dimensions 6 m by 1 m by m, and so has volume 6 1 = 144 m 3. Answer: (E) 13. To determine the number that must be added 8 to give the result 5, we subtract 8 from 5 to get ( 5) 8 = 13. Checking, 8 + ( 13) = 5. Answer: (D) Since AOB is a diameter of the circle, then AOB = 180. We are told that the angle in the Winter sector is a right angle (or 90 ). Also, we are told that the angle in the Spring sector is 60. Therefore, the angle in the Fall sector is = 30. What fraction of the complete circle is 30? Since the whole circle has 360, then the fraction is = Therefore, of the students chose fall as their favourite season, or 600 = 50 students in 1 1 total. Since AOB is a diameter of the circle, then 1 of the students chose summer as their favourite season, or = 300 students in total. Since the angle in the Winter sector is a right angle (or 90 ), then 1 of the students (since 4 4 right angles make up a complete circle) chose Winter as their favourite season, or = students in total. Since the angle in the Spring sector is 60, then = 1 of the students chose Spring as 6 their favourite season, or = 100 students in total. 6 Since there were 600 students in total, then the number who chose Fall as their favourite season was = 50. Answer: (B) 15. Since Harry charges 50% more for each additional hour as he did for the previous hour, then he charges 1.5 or 3 times as much as he did for the previous hour. Harry charges $4 for the first hour. Harry then charges 3 $4 = $6 for the second hour. Harry then charges 3 $6 = $9 for the third hour. Harry then charges 3 $9 = $ 7 = $13.50 for the fourth hour. Therefore, for 4 hours of babysitting, Harry would earn $4 + $6 + $9 + $13.50 = $3.50. Answer: (C)

144 006 Gauss Contest s Page We obtain fractions equivalent to 5 by multiplying the numerator and denominator by the same 8 number. The sum of the numerator and denominator of 5 is 13, so when we multiply the numerator and 8 denominator by the same number, the sum of the numerator and denominator is also multiplied by this same number. Since 91 = 13 7, then we should multiply the numerator and denominator both by 7 to get a fraction 5 7 = 35 equivalent to 5 whose numerator and denominator add up to The difference between the denominator and numerator in this fraction is = 1. We make a list of the fractions equivalent to 5 by multiplying the numerator and denominator 8 by the same number, namely, 3, 4, and so on: 5, 10, 15, 0, 5, 30, 35, Since the numerator and denominator of 35 add to 91 (since = 91), then this is the 56 fraction for which we are looking. The difference between the denominator and numerator is = 1. 3 We obtain fractions equivalent to 5 by multiplying the numerator and denominator by the same 8 number. If this number is n, then a fraction equivalent to 5 5n is. 8 8n For the numerator and denominator to add up to 91, we must have 5n + 8n = 91 or 13n = 91 or n = 7. Therefore, the fraction for which we are looking is 5 7 = The difference between the denominator and numerator is = 1. Answer: (A) Since the shoe is 8 cm long and fits 15 times along one edge of the carpet, then one dimension of the carpet is 15 8 = 40 cm. Since the shoe fits 10 times along another edge of the carpet, then one dimension of the carpet is 10 8 = 80 cm. Therefore, then area of the carpet is = cm. Since the shoe fits along one edge of the carpet 15 times and along another edge 10 times, then the area of the carpet is = 150 square shoes. Since the length of the shoe is 8 cm, then 1 square shoe = 1 shoe 1 shoe = 8 cm 8 cm = 784 cm Therefore, the area of the carpet in cm is = cm. Answer: (E) Since Keiko takes 10 seconds to run 3 times around the track, then it takes her 10 = 40 seconds to run 1 time around the track. 1 3

145 006 Gauss Contest s Page 6 Since Leah takes 160 seconds to run 5 times around the track, then it takes her = 3 seconds to run 1 time around the track. 5 Since Leah takes less time to run around the track than Keiko, then she is the faster runner. Since Leah takes 3 seconds to run the 150 m around the track, then her speed is 150 m = m/s 4.69 m/s. 3 s Therefore, Leah is the faster runner and her speed is approximately 4.69 m/s. In 10 seconds, Keiko runs 3 times around the track, or = 450 m in total. Therefore, her speed is 450 m = 3.75 m/s. 10 s In 160 seconds, Leah runs 5 times around the track, or = 750 m in total. Therefore, her speed is 750 m = m/s 4.69 m/s. 160 s Since Leah s speed is larger, she is the faster runner and her speed is approximately 4.69 m/s. Answer: (D) In one minute, there are 60 seconds. In one hour, there are 60 minutes, so there are = 3600 seconds. In one day, there are 4 hours, so there are = seconds. Therefore, seconds is equal to days, which of the given choices is closest to 10 days. Since there are 60 seconds in one minute, then 10 6 seconds is Since there are 60 minutes in one hour, then minutes is Since there are 4 hours in one day, then hours is given choices is closest to 10 days minutes hours days, which of the Answer: (B) 0. One possible way to transform the initial position of the P to the final position of the P is to reflect the grid in the vertical line in the middle to obtain and then rotate the grid 90 counterclockwise about the centre to obtain. Applying these transformations to the grid containing the A, we obtain and then. (There are many other possible combinations of transformations which will produce the same resulting image with the P ; each of these combinations will produce the same result with the A.) Answer: (B)

146 006 Gauss Contest s Page Between x a.m. and x p.m. there are 1 hours. (For example, between 10 a.m. and 10 p.m. there are 1 hours.) Therefore, Gail works for 1 hours on Saturday. From x a.m. until 1 noon, the number of hours which Gail works is 1 x. From 1 noon until x p.m., she works x hours. Thus, the total number of hours that Gail works is (1 x) + x = 1. Answer: (E). As an initial guess, let us see what happens when the line passes through C. P X Z Y A B E D C Since each square is a unit square, then the area of rectangle P XCY is 3 = 6, and so the line through P and C cuts this area in half, leaving 3 square units in the bottom piece and 3 square units in the top piece. (A fact has been used here that will be used several times in this solution: If a line passes through two diagonally opposite vertices of a rectangle, then it cuts the rectangle into two pieces of equal area (since it cuts the rectangle into two congruent triangles).) In the bottom piece, we have not accounted for the very bottom unit square, so the total area of the bottom piece is 4 square units. In the top piece, we have not accounted for the two rightmost unit squares, so the total area of the top piece is 5 square units. So putting the line through C does not produce two pieces of equal area, so C is not the correct answer. Also, since the area of the bottom piece is larger than the area of the top piece when the line passes through C, we must move the line up to make the areas equal (so neither A nor B can be the answer).

147 006 Gauss Contest s Page 8 Should the line pass through E? P X Z Y A B E D C If so, then the line splits the square P XEZ (of area 4) into two pieces of area. Then accounting for the remaining squares, the area of the bottom piece is + 3 = 5 and the area of the top piece is + = 4. So putting the line through E does not produce two pieces of equal area, so E is not the correct answer. By elimination, the correct answer should be D. (We should verify that putting the line through D does indeed split the area in half. The total area of the shape is 9, since it is made up of 9 unit squares. If the line goes through D, the top piece consists of P XD and unit squares. The area of P XD is 1 5 = 5, since P X = and XD = 5. Thus, the area of the top piece is 5 + = 9, which is exactly half of the total area, as required.) Answer: (D) 3. We label the blank spaces to make them easier to refer to. + A C B D E F G Since we are adding two -digit numbers, then their sum cannot be 00 or greater, so E must be a 1 if the sum is to have 3 digits. Where can the digit 0 go? Since no number can begin with a 0, then neither A nor C can be 0. Since each digit is different, then neither B and D can be 0, otherwise both D and G or B and G would be the same. Therefore, only F or G could be 0. Since we are adding two -digit numbers and getting a number which is at least 100, then A+C must be at least 9. (It could be 9 if there was a carry from the sum of the units digits.) This tells us that A and C must be 3 and 6, 4 and 5, 4 and 6, or 5 and 6. If G was 0, then B and D would have to 4 and 6 in some order. But then the largest that A and C could be would be 3 and 5, which are not among the possibilities above. Therefore, G is not 0, so F = 0.

148 006 Gauss Contest s Page 9 + A C B D 1 0 G So the sum of A and C is either 9 or 10, so A and C are 3 and 6, 4 and 5, or 4 and 6. In any of these cases, the remaining possibilities for B and D are too small to give a carry from the units column to the tens column. So in fact, A and C must add to 10, so A and C are 4 and 6 in some order. Let s try A = 4 and C = B 6 D 1 0 G The remaining digits are, 3 and 5. To make the addition work, B and D must be and 3 and G must be 5. (We can check that either order for B and D works, and that switching the 4 and 6 will also work.) So the units digit of the sum must be 5, as in the example (Note that we could have come up with this answer by trial and error instead of this logical procedure.) Answer: (D) 4. The sum of any two sides of a triangle must be bigger than the third side. (When two sides are known to be equal, we only need to check if the sum of the two equal sides is longer than the third side, since the sum of one of the equal sides and the third side will always be longer than the other equal side.) If the equal sides were both equal to, the third side must be shorter than + = 4. The 1 possibility from the list not equal to (since we cannot have three equal sides) is 3. So here there is 1 possibility. If the equal sides were both equal to 3, the third side must be shorter than = 6. The possibilities from the list not equal to 3 (since we cannot have three equal sides) are and 5. So here there are possibilities. If the equal sides were both equal to 5, the third side must be shorter than = 10. The 3 possibilities from the list not equal to 5 (since we cannot have three equal sides) are, 3 and 7. So here there are 3 possibilities. If the equal sides were both equal to 7, the third side must be shorter than = 14. The 4 possibilities from the list not equal to 7 (since we cannot have three equal sides) are, 3, 5

149 006 Gauss Contest s Page 10 and 11. So here there are 4 possibilities. If the equal sides were both equal to 11, the third side must be shorter than =. The 4 possibilities from the list not equal to 11 (since we cannot have three equal sides) are, 3, 5 and 7. So here there are 4 possibilities. Thus, in total there are = 14 possibilities. Answer: (E) 5. The five scores are N, 4, 43, 46, and 49. If N < 43, the median score is 43. If N > 46, the median score is 46. If N 43 and N 46, then N is the median. We try each case. If N < 43, then the median is 43, so the mean should be 43. Since the mean is 43, then the sum of the 5 scores must be 5 43 = 15. Therefore, N = 15 or N = 15 or N = 35, which is indeed less than 43. We can check that the median and mean of 35, 4, 43, 46 and 49 are both 43. If N > 46, then the median is 46, so the mean should be 46. Since the mean is 46, then the sum of the 5 scores must be 5 46 = 30. Therefore, N = 30 or N = 30 or N = 50, which is indeed greater than 46. We can check that the median and mean of 4, 43, 46, 49, and 50 are both 46. If N 43 and N 46, then the median is N, so the mean should be N. Since the mean is N, then the sum of the 5 scores must be 5N. Therefore, N = 5N or N = 5N or 4N = 180, or N = 45, which is indeed between 43 and 46. We can check that the median and mean of 4, 43, 45, 46 and 49 are both 45. Therefore, there are 3 possible values for N. Answer: (A)

150 006 Gauss Contest s Page 11 Grade 8 1. Calculating using the correct order of operations, 30 5 = 30 5 = 5. Answer: (E). 1 Since 98 = 49, 98 7 = 14, = 7, =, and 98 4 = 4.5, then the one of the five choices which does not divide exactly into 98 is 4. The prime factorization of 98 is 98 = 7 7. Of the given possibilities, only 4 = cannot be a divisor of 98, since there are not two s in the prime factorization of 98. Answer: (B) 3. Since the tax is 15% on the $00.00 camera, then the tax is 0.15 $00.00 = $ Answer: (A) 4. Since = 3.1, then to get a sum of 4.44, we still need to add = 1.3. Thus, the number which should be put in the box is 1.3. Answer: (B) 5. In total, there are = 11 balls in the bag. Since there are 5 yellow balls, then the probability of choosing a yellow ball is 5 11 Ȧnswer: (B) 6. We check each number between 0 and 30. None of 0,, 4, 6, 8, and 30 is a prime number, because each has a factor of. Neither 1 nor 7 is a prime number, since each has a factor of 3. Also, 5 is not a prime number, since it has a factor of 5. We do not need to look for any larger prime factors, since this leaves only 3 and 9, each of which is a prime number. Therefore, there are prime numbers between 0 and 30. Answer: (C) 7. Since the volume of a rectangular block is equal to the area of its base times its height, then 10 cm3 the height of this particular rectangular block is 4 cm = 5 cm. Answer: (A) 8. Since the rate of rotation of the fan doubles between the slow and medium settings and the fan rotates 100 times in 1 minute on the slow setting, then it rotates 100 = 00 times in 1 minute on the medium setting. Since the rate of rotation of the fan doubles between the medium and high setting, then it rotates 00 = 400 times in 1 minute on the high setting. Therefore, in 15 minutes, it will rotate = 6000 times. Answer: (C)

151 006 Gauss Contest s Page A 60 X 50 B C 10 Y x Z D Since AXB = 180, then Y XZ = = 70. Also, XY Z = 180 CY X = = 60. Since the angles in XY Z add to 180, then x = = 50, so x = 50. Since CY X + AXY = 180, then AB is parallel to CD. Therefore, Y ZX = ZXB or x = 50 or x = 50. Answer: (A) When we divide 836 by 1, we obtain so = In other words, 836 lollipops make up 696 complete packages leaving 10 1 lollipops left over. When we divide 836 by 1, we obtain approximately , so the maximum possible number of packages which can be filled is 696. In total, 696 packages contain 835 lollipops, leaving 10 remaining from the initial 836 lollipops. Answer: (E) 11. Since the sound of thunder travels at 331 m/s and the thunder is heard 1 seconds after the lightning flash, then Joe is 1 s 331 m/s = 397 m = 3.97 km from the lightning flash, or, to the nearest tenth of a kilometre, 4.0 km. Answer: (C) 1. The shaded triangle has a base of length 10 cm. Since the triangle is enclosed in a rectangle of height 3 cm, then the height of the triangle is 3 cm. (We know that the enclosing shape is a rectangle, because any figure with 4 sides, including pairs of equal opposite sides, and right angles must be a rectangle.) Therefore, the area of the triangle is = 15 cm. Answer: (C)

152 006 Gauss Contest s Page We need to find two consecutive numbers the first of which is a multiple of 7 and the second of which is a multiple of 5. We try the multiples of 7 and the numbers after each. Do 7 and 8 work? No, since 8 is not a multiple of 5. Do 14 and 15 work? Yes, since 15 is a multiple of 5. This means that this year, Kiril is 15 years old. So it will be 11 years until Kiril is 6 years old. (Of course, Kiril could also be 50 years old or 85 years old this year, but then he would have already been 6 years old in the past.) Answer: (A) 14. We are told that the second term in the sequence is 60. Using the rule for the sequence, to get the third term, we divide the second term by to obtain 130 and then add 10 to get 140, which is the third term. To get the fourth term, we divide the third term by to obtain 70 and then add 10 to get 80. Answer: (E) 15. When the original shape F is reflected in Line 1 (that is, reflected vertically), the shape is obtained. When this new shape is reflected in Line (that is, reflected horizontally), the shape that results is F. F Answer: (D) 16. By the Pythagorean Theorem in ABD, we have BD + 16 = 0 or BD + 56 = 400 or BD = 144. Therefore, BD = 1. A 16 B 1 0 D 5 C By the Pythagorean Theorem in BDC, we have BC = = = 169, so BC = 169 or BC = 13. Answer: (A) 17. Since 10 x 10 = 9990, then 10 x = = If 10 x = , then x = 4, since ends in 4 zeroes. Answer: (D) 18. Since the square has perimeter 4, then the side length of the square is 1 4 = 6. 4 Since the square has side length 6, then the area of the square is 6 = 36. Since the rectangle and the square have the same area, then the area of rectangle is 36. Since the rectangle has area 36 and width 4, then the length of the rectangle is 36 4 = 9. Since the rectangle has width 4 and length 9, then it has perimeter = 6. Answer: (A)

153 006 Gauss Contest s Page We can write out the possible arrangements, using the first initials of the four people, and remembering that there two different ways in which Dominic and Emily can sit side by side: DEBC, DECB, EDBC, EDCB, BDEC, CDEB, BEDC, CEDB, BCDE, CBDE, BCED, CBED There are 1 possible arrangements. Since Dominic and Emily must sit beside other, then we can combine then into one person either Domily or Eminic, depending on the order in which they sit and their two seats into one seat. We now must determine the number of ways of arranging Bethany, Chun and Domily into three seats, and the number of ways of arranging Bethany, Chun and Eminic into three seats. With three people, there are three possible choices for who sits in the first seat, and for each of these choices there are two possible choices for the next seat, leaving the choice for the final seat fixed. So with three people, there are 3 = 6 possible ways that they can sit in three chairs. So for each of the ways in which Dominic and Emily can sit, there are 6 ways that the seating can be done, for a total of 1 seating arrangements. Answer: (C) 0. We label the blank spaces to make them easier to refer to. + A C B D E F G Since we are adding two -digit numbers, then their sum cannot be 00 or greater, so E must be a 1 if the sum is to have 3 digits. Where can the digit 0 go? Since no number can begin with a 0, then neither A nor C can be 0. Since each digit is different, then neither B and D can be 0, otherwise both D and G or B and G would be the same. Therefore, only F or G could be 0. Since we are adding two -digit numbers and getting a number which is at least 100, then A+C must be at least 9. (It could be 9 if there was a carry from the sum of the units digits.) This tells us that A and C must be 3 and 6, 4 and 5, 4 and 6, or 5 and 6. If G was 0, then B and D would have to 4 and 6 in some order. But then the largest that A and C could be would be 3 and 5, which are not among the possibilities above. Therefore, G is not 0, so F = 0.

154 006 Gauss Contest s Page 15 + A C B D 1 0 G So the sum of A and C is either 9 or 10, so A and C are 3 and 6, 4 and 5, or 4 and 6. In any of these cases, the remaining possibilities for B and D are too small to give a carry from the units column to the tens column. So in fact, A and C must add to 10, so A and C are 4 and 6 in some order. Let s try A = 4 and C = B 6 D 1 0 G The remaining digits are, 3 and 5. To make the addition work, B and D must be and 3 and G must be 5. (We can check that either order for B and D works, and that switching the 4 and 6 will also work.) So the units digit of the sum must be 5, as in the example (Note that we could have come up with this answer by trial and error instead of this logical procedure.) Answer: (D) 1. 1 If Nathalie had exactly 9 quarters, 3 dimes and 1 nickel, she would have = 60 cents or $.60 in total. Since the coins she has are in this same ratio 9 : 3 : 1, then we can split her coins up into sets of 9 quarters, 3 dimes and 1 nickel, with each set having a value of $.60. Since the total value of her coins is $18.0 and $18.0 = 7, then she has 7 of these sets of coins. $.60 Since each of these sets contains 13 coins, then she has 7 13 = 91 coins in total. Suppose Nathalie had n nickels. Since the ratio of the number of quarters to the number of dimes to the number of nickels that she has is 9 : 3 : 1, then she must have 3n dimes and 9n quarters. This tells us that the total value of her coins in cents is (9n 5) + (3n 10) + (n 5) = 60n. Since we know that the total value of her coins is 180 cents, then 60n = 180 or n = 7. Therefore, she has 7 nickels, 1 dimes and 63 quarters, or 91 coins in total. Answer: (D)

155 006 Gauss Contest s Page 16. Label the first 8 people at the party as A, B, C, D, E, F, G, and H. Then A shakes hands with the 7 others, B accounts for 6 more handshakes (since we have already included B shaking hands with A), C accounts for 5 more handshakes (since we have already included C shaking hands with A and B), D accounts for 4 more handshakes, E accounts for 3 more handshakes, F accounts for more handshakes, and G accounts for 1 more handshake. (H s handshakes with each of the others have already been included.) So there are = 8 handshakes which take place before the ninth person arrives. Since there are a total of 3 handshakes which take place, then the ninth person shakes hands with 3 8 = 4 people. Answer: (B) 3. In how many ways can we place C so that AB = AC? To get from A to B we go 1 unit in one direction and units in the perpendicular direction from A. To choose C so that AB = AC, we must also go 1 unit in one direction and units in the perpendicular direction. This gives the 3 possible points: A B In how many ways can we place C so that BA = BC? To choose C so that BA = BC, we again also go 1 unit in one direction and units in the other direction from B. This gives the 3 possible points: A B So in total we have 6 possible points so far, which is the largest of the possible answers. So 6 must be the answer. (There is one more possibility: can C be placed so that CA = CB? Since 6 must be the answer, then the answer to this question is no. We can also check this by trying to find such points in the diagram.) Answer: (A) 4. In the 1st row, every box is shaded. In the nd row, the boxes whose column numbers are multiples of are shaded. In the 3rd row, the boxes whose column numbers are multiples of 3 are shaded. In the nth row, the boxes whose column numbers are multiples of n are shaded. So in a particular column, the boxes which are shaded are those which belong to row numbers which are divisors of the column number. (We can see this for instance in columns 4 and 6 where the boxes in rows 1, and 4, and 1,,

156 006 Gauss Contest s Page 17 3, and 6, respectively, are shaded.) So to determine which of the given columns has the largest number of shaded boxes, we must determine which of the given numbers has the greatest number of divisors. To find the divisors of 144, for instance, it is easier to find the prime factors of 144 first. To do this, we see that 144 = 16 9 = 3 3 = 4 3. We can then write out the divisors of 144, which are 1,, 3, 4, 6, 8, 9, 1, 16, 18, 4, 36, 48, 7, 144, or a total of 15 divisors. Similarly, 10 = 3 3 5, so we can find the divisors of 10, which are 1,, 3, 4, 5, 6, 8, 10, 1, 15, 0, 4, 30, 40, 60, 10, or a total of 16 divisors. Since 150 = 3 5, the divisors of 150 are 1,, 3, 5, 6, 10, 15, 5, 30, 50, 75, 150, or a total of 1 divisors. Since 96 = 5 3, the divisors of 96 are 1,, 3, 4, 6, 8, 1, 16, 4, 3, 48, 96, or a total of 1 divisors. Since 100 = 5, the divisors of 100 are 1,, 4, 5, 10, 0, 5, 50, 100, or a total of 9 divisors. So the number with the most divisors (and thus the column with the most shaded boxes) is 10. Answer: (B) 5. The numbers which have yet to be placed in the grid are 10, 11, 1, 13, 14, 15, 16, 17, 18, and 19. We look at the numbers in the corners first, because they have the fewest neighbours. Look first at the 5 in the bottom left corner. It must be the sum of two of its neighbours, so it must be or Since the 1 has already been placed in the grid, then the empty space next to the 5 must be filled with the 16. Look next at the 1 in the top right corner. It must be equal to (which it cannot be since the 1 has already been placed) or , so the 17 must be placed in the space next to the ? Since the 17 must be the sum of two of its neighbours, then the 17 must be or , so the? must be replaced by either the 13 or the 14. Consider the in the corner. Since the 0 is already placed, cannot be + 0. So the is the sum of its two missing neighbours. Since the possible missing neighbours are 10, 11, 1, 13, 14, 15, and 18, then the two neighbours must be 10 and 1 in some order. However, the 10 cannot go above the, since there it could not be the sum of two of its neighbours (since the 7 and 8 have already been placed). Therefore, the 1 goes above the, so we have

157 006 Gauss Contest s Page ? The 13 now cannot replace the? since 13 is not the sum of any two of 17, 4, 3,, and 1. Therefore, the? is replaced by the 14. At this stage, we are already finished, but we could check that the grid does complete as follows: Answer: (C)

158 Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 005 Gauss Contests (Grades 7 and 8) Wednesday, May 11, 005 s c 004 Waterloo Mathematics Foundation

159 005 Gauss Contest s Page 3 Grade 7 1. Calculating the numerator first, = 1 6 =. Answer: (B). Calculating, = = Answer: (E) 3. Since the arrow is pointing between 9.6 and 9.8, it is pointing to a rating closest to 9.7. Answer: (C) 4. Twelve million is written as and twelve thousand is written as 1 000, so the sum of these two numbers is Answer: (A) 5. To figure out which number is largest, we look first at the number in the tenths position. Since four of the given numbers have a 1 in the tenths position and 0. has a, then 0. is the largest. Answer: (B) 6. Since Meghan chooses a prize from 7 in the bag and the probability of her choosing a book is, then of the prizes in the bag must be books. 3 3 Therefore, the number of books in the bag is 7 = Answer: (E) 7. Since 83% in decimal form is 0.83, then the number of people who voted for Karen is equal to = Answer: (B) 8. Since ABC + ABD = 180 (in other words, ABC and ABD are supplementary) and ABD = 130, then ABC = 50. A 93 o D 130 o 50 o B C Since the sum of the angles in triangle ABC is 180 and we know two angles 93 and 50 which add to 143, then ACB = = 37. Answer: (B) 9. There are six odd-numbered rows (rows 1, 3, 5, 7, 9, 11). These rows have 6 15 = 90 seats in total. There are five even-numbered rows (rows, 4, 6, 8, 10). These rows have 5 16 = 80 seats in total. Therefore, there are = 170 seats in total in the theatre. Answer: (D)

160 005 Gauss Contest s Page When it is 5:36 p.m. in St. John s, it is 90 minutes or 1 1 hours earlier in Smiths Falls, so it is 4:06 p.m. in Smiths Falls. When it is 4:06 p.m. in Smiths Falls, it is 3 hours earlier in Whitehorse, so it is 1:06 p.m. in Whitehorse. Answer: (A) 11. On each day, the temperature range is the difference between the daily high and daily low temperatures. On Monday, the range is 6 ( 4) = 10 degrees Celsius. On Tuesday, the range is 3 ( 6) = 9 degrees Celsius. On Wednesday, the range is 4 ( ) = 6 degrees Celsius. On Thursday, the range is 4 ( 5) = 9 degrees Celsius. On Friday, the range is 8 0 = 8 degrees Celsius. The day with the greatest range is Monday. Answer: (A) 1. Since the bamboo plant grows at a rate of 105 cm per day and there are 7 days from May 1st and May 8th, then it grows = 735 cm in this time period. Since 735 cm = 7.35 m, then the height of the plant on May 8th is = 9.35 m. Answer: (E) 13. Since BD = 3 and DC is twice the length of BD, then DC = 6. A B D C Therefore, triangle ABC has a base of length 9 and a height of length 4. Therefore, the area of triangle ABC is 1bh = 1(9)(4) = 1 (36) = 18. Answer: (D) Since the sum of the numbers on opposite faces on a die is 7, then 1 and 6 are on opposite faces, and 5 are on opposite faces, and 3 and 4 are on opposite faces. On the first die, the numbers on the unseen faces opposite the 6, and 3 are 1, 5 and 4, respectively. On the second die, the numbers on the unseen faces opposite the 1, 4 and 5 are 6, and 3, respectively. The sum of the missing numbers is = 1. The sum of the numbers on a die is = 1 and so the sum of the numbers on two die is 1 = 4. Since there is a sum of 1 showing on the six visible faces, the sum of the numbers on the six unseen faces is 4 1 = 1. Answer: (C)

161 005 Gauss Contest s Page Since the area of rectangle P QRS is 4, let us assume that P Q = 6 and QR = 4. Since QT = QR, then QR = QT so QT =. P S 6 Q T R Therefore, triangle P QT has base P Q of length 6 and height QT of length, so has area 1 (6)() = 1 (1) = 6. So the area of quadrilateral P T RS is equal to the area of rectangle P QRS (which is 4) minus the area of triangle P QT (which is 6), or 18. Draw a line through T parallel to P Q across the rectangle parallel so that it cuts P S at point V. P V S Q T R Since T is halfway between Q and R, then V is halfway between P and S. Therefore, SV T R is a rectangle which has area equal to half the area of rectangle P QRS, or 1. Similarly, V P QT is a rectangle of area 1, and V P QT is cut in half by P T, so triangle P V T has area 6. Therefore, the area of P T RS is equal to the sum of the area of rectangle SV T R and the area of triangle P V T, or = 18. Answer: (A) 16. Nicholas sleeps for an hour and a half, or 90 minutes. Since three sleep cross the road per minute, then 3 90 = 70 sheep cross while he is asleep. Since 4 sheep crossed before he fell asleep, then = 31 sleep have crossed the road in total when he wakes up. Since this is half of the total number of sheep in the flock, then the total number in the flock is 31 = 64. Answer: (D) When we calculate the value of the symbol, we add the product of the numbers on each of the two diagonals. The product of the entries on the diagonal with the 1 and the 6 is 6. Since the symbol is evaluated as 16, then the product of the entries on the other diagonal is 10. Since one of the entries on the other diagonal is, then the missing entry must be 5.

162 005 Gauss Contest s Page 6 Let the missing number be x. Using the definition for the evaluation of the symbol, we know that x = 16 or x + 6 = 16 or x = 10 or x = 5. Answer: (E) 18. When a die is rolled, there are six equally likely possibilities (1 through 6). In order for the game to be fair, half of the six possibilities, or three possibilities, must be winning possibilities. In the first game, only rolling a gives a win, so this game is not fair. In the second game, rolling a, 4 or 6 gives a win, so this game is fair. In the third game, rolling a 1, or 3 gives a win, so this game is fair. In the fourth game, rolling a 3 or 6 gives a win, so this game is not fair. Therefore, only two of the four games are fair. Answer: (C) 19. At each distance, two throws are made: the 1st and nd throws are made at 1 m, the 3rd and 4th are made at m, and so on, with the 7th and 8th throws being made at 14 m. Therefore, the 9th throw is the first throw made at 15 m. At each distance, the first throw is made by Pat to Chris, so Chris misses catching the 9th throw at a distance of 15 m. Answer: (A) 0. Since Sally s car travels 80 km/h, it travels m in one hour. Since there are 60 minutes in an hour, the car travels m in one minute. 60 Since there are 60 seconds in a minute, the car travels in one second Therefore, in 4 seconds, the car travels m Of the possible choices, this is closest to 90 m. Answer: (E) 1. Since the price of the carpet is reduced by 10% every 15 minutes, then the price is multiplied by 0.9 every 15 minutes. At 9:15, the price was $9.00. At 9:30, the price fell to 0.9 $9.00 = $8.10. At 9:45, the price fell to 0.9 $8.10 = $7.9. So the price fell below $8.00 at 9:45 a.m., so Emily bought the carpet at 9:45 a.m. Answer: (A). 1 We start by assuming that there are 0 oranges. (We pick 0 since the ratio of apples to oranges is 1 : 4 and the ratio of oranges to lemons is 5 :, so we pick a number of oranges which is divisible by 4 and by 5. Note that we did not have to assume that there were 0 oranges, but making this assumption makes the calculations much easier.) Since there are 0 oranges and the ratio of the number of apples to the number of oranges is 1 : 4, then there are 1 0 = 5 apples. 4 Since there are 0 oranges and the ratio of the number of oranges to the number of lemons is 5 :, then there are 0 = 8 lemons. 5 Therefore, the ratio of the number of apples to the number of lemons is 5 : 8.

163 005 Gauss Contest s Page 7 Let the number of apples be x. Since the ratio of the number of apples to the number of oranges is 1 : 4, then the number of oranges is 4x. Since the ratio of the number of oranges to the number of lemons is 5 :, then the number of lemons is 4x = 8x. 5 5 Since the number of apples is x and the number of lemons is 8 x, then the ratio of the number 5 of apples to the number of lemons is 1 : 8 = 5 : 8. 5 Answer: (C) 3. 1 If 4 balance, then 1 would balance the equivalent of 1. Similarly, 1 would balance the equivalent of 1 1. If we take each of the answers and convert them to an equivalent number of (A): = 3 (B): 3 ( ) = 3 (C): ( 1 ) + = 3 (D): ( 1 1 ) + 1 = 3 1 (E): ( 1 ) = 3 Therefore, and 1 do not balance the required., we would have: Since 4 balance, then 1 would balance. Therefore, 3 would balance 6, so since 3 balance, then 6 would balance, or 1 would balance 3. We can now express every combination in terms of only. 1, 1 and 1 equals = 6. 3 and 1 equals = 6. and equals + = 6. and 1 equals = 7. 1 and 4 equals + 4 = 6. Therefore, since 1, 1 and 1 equals 6, then it is and 1 which will not balance with this combination. 3 We try assigning weights to the different shapes. Since 3 balance, assume that each weighs kg and each weighs 3 kg. Therefore, since 4 balance, which weigh 4 kg combined, then each weighs 1 kg. We then look at each of the remaining combinations. 1, 1 and 1 weigh = 6 kg. 3 and 1 weigh = 6 kg. and weigh + = 6 kg. and 1 weigh = 7 kg. 1 and 4 weigh + 4 = 6 kg. Therefore, it is the combination of and 1 which will not balance the other combinations. Answer: (D)

164 005 Gauss Contest s Page 8 4. Since Alphonse and Beryl always pass each other at the same three places on the track and since they each run at a constant speed, then the three places where they pass must be equally spaced on the track. In other words, the three places divide the track into three equal parts. We are not told which runner is faster, so we can assume that Beryl is the faster runner. Start at one place where Alphonse and Beryl meet. (Now that we know the relative positions of where they meet, we do not actually have to know where they started at the very beginning.) To get to their next meeting place, Beryl runs farther than Alphonse (since she runs faster than he does), so Beryl must run of the track while Alphonse runs 1 of the track in the opposite 3 3 direction, since the meeting placed are spaced equally at 1 intervals of the track. 3 Since Beryl runs twice as far in the same length of time, then the ratio of their speeds is : 1. Answer: (D) 5. 1 We want to combine 48 coins to get 100 cents. Since the combined value of the coins is a multiple of 5, as is the value of a combination of nickels, dimes and quarters, then the value of the pennies must also be a multiple of 5. Therefore, the possible numbers of pennies are 5, 10, 15, 0, 5, 30, 35, 40. We can also see that because there are 48 coins in total, it is not possible to have anything other than 35, 40 or 45 pennies. (For example, if we had 30 pennies, we would have 18 other coins which are worth at least 5 cents each, so we would have at least = 10 cents in total, which is not possible. We can make a similar argument for 5, 10, 15, 0 and 5 pennies.) It is also not possible to have 3 or 4 quarters. If we did have 3 or 4 quarters, then the remaining 45 or 44 coins would give us a total value of at least 44 cents, so the total value would be greater than 100 cents. Therefore, we only need to consider 0, 1 or quarters. Possibility 1: quarters If we have quarters, this means we have 46 coins with a value of 50 cents. The only possibility for these coins is 45 pennies and 1 nickel. Possibility : 1 quarter If we have 1 quarter, this means we have 47 coins with a value of 75 cents. The only possibility for these coins is 40 pennies and 7 nickels. Possibility 3: 0 quarters If we have 0 quarters, this means we have 48 coins with a value of 100 cents. If we had 35 pennies, we would have to have 13 nickels. If we had 40 pennies, we would have to have 4 dimes and 4 nickels. It is not possible to have 45 pennies. Therefore, there are 4 possible combinations. We want to use 48 coins to total 100 cents. Let us focus on the number of pennies. Since any combination of nickels, dimes and quarters always is worth a number of cents which is divisible by 5, then the number of pennies in each combination must be divisible by 5, since the total value of each combination is 100 cents, which is divisible by 5.

165 005 Gauss Contest s Page 9 Could there be 5 pennies? If so, then the remaining 43 coins are worth 95 cents. But each of the remaining coins is worth at least 5 cents, so these 43 coins are worth at least 5 43 = 15 cents, which is impossible. So there cannot be 5 pennies. Could there be 10 pennies? If so, then the remaining 38 coins are worth 90 cents. But each of the remaining coins is worth at least 5 cents, so these 38 coins are worth at least 5 38 = 190 cents, which is impossible. So there cannot be 10 pennies. We can continue in this way to show that there cannot be 15, 0, 5, or 30 pennies. Therefore, there could only be 35, 40 or 45 pennies. If there are 35 pennies, then the remaining 13 coins are worth 65 cents. Since each of the remaining coins is worth at least 5 cents, this is possible only if each of the 13 coins is a nickel. So one combination that works is 35 pennies and 13 nickels. If there are 40 pennies, then the remaining 8 coins are worth 60 cents. We now look at the number of quarters in this combination. If there are 0 quarters, then we must have 8 nickels and dimes totalling 60 cents. If all of the 8 coins were nickels, they would be worth 40 cents, so we need to change 4 nickels to dimes to increase our total by 0 cents to 60 cents. Therefore, 40 pennies, 0 quarters, 4 nickels and 4 dimes works. If there is 1 quarter, then we must have 7 nickels and dimes totalling 35 cents. Since each remaining coin is worth at least 5 cents, then all of the 7 remaining coins must be nickels. Therefore, 40 pennies, 1 quarter, 7 nickels and 0 dimes works. If there are quarters, then we must have 6 nickels and dimes totalling 10 cents. This is impossible. If there were more than quarters, the quarters would be worth more than 60 cents, so this is not possible. If there are 45 pennies, then the remaining 3 coins are worth 55 cents in total. In order for this to be possible, there must be quarters (otherwise the maximum value of the 3 coins would be with 1 quarter and dimes, or 45 cents). This means that the remaining coin is worth 5 cents, and so is a nickel. Therefore, 45 pennies, quarters, 1 nickel and 0 dimes is a combination that works. Therefore, there are 4 combinations that work. Answer: (B)

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167 005 Gauss Contest s Page 11 Grade 8 1. Using a common deminator of 8, = = 5 8. Answer: (B). Calculating, ( 3)( 4)( 1) = (1)( 1) = Since V = l w h, the required volume is 4 8 = 64 cm 3. Answer: (A) Answer: (C) 4. The mean of these five numbers is = 50 5 = 10. Answer: (C) 5. 10% of 10 is = 1 or 1 10 = % of 0 is 0. 0 = 4 or 1 0 = 4. 5 Therefore, 10% of 10 times 0% of 0 equals 1 4 = = so we must have 10 = 1000 so the required number is 3. Answer: (E) Answer: (C) 7. Since ABC+ BAC+ BCA = 180 and ABC = 80 and BAC = 60, then BCA = 40. B A 60 o 80 o 40 o C y o D E Since DCE = BCA = 40, and looking at triangle CDE, we see that DCE+ CED = 90 then 40 + y = 90 or y = 50. Answer: (D) 8. 1 There are 10 numbers (30 to 39) which have a tens digit of 3. There are 6 numbers (3, 13, 3, 33, 43, 53) which have a units digit of 3. In these two lists, there is one number counted twice, namely 33. Therefore, the total number of different numbers in these two lists is = 15. We list out the numbers in increasing order: 3, 13, 3, 30, 31, 3, 33, 34, 35, 36, 37, 38, 39, 43, 53. There are 15 of these numbers. Answer: (D)

168 005 Gauss Contest s Page 1 9. Since the average monthly rainfall was 41.5 mm in 003, then the average monthly rainfall in 004 was 41 + = 43.5 mm. Therefore, the total rainfall in 004 was = 5 mm. Answer: (B) 10. Since Daniel rides at a constant speed, then, in 30 minutes, he rides 3 4 does in 40 minutes. Therefore, in 30 minutes, he rides 3 4 = 18 km. 4 Answer: of the distance that he (D) 11. Triangle ABC has base AB of length 5 cm and height AC of length 0 cm. Therefore, the area of triangle ABC is 1 bh = 1 (5 cm)(0 cm) = 1 (500 cm ) = 50 cm. Answer: (E) 1. To make the sum of the five consecutive even numbers including 10 and 1 as large as possible, we should make 10 and 1 the smallest of these five numbers. Therefore, to make the sum as large as possible, the numbers should be 10, 1, 14, 16, and 18, which have a sum of 70. Answer: (E) 13. Since the sum of the angles at any point on a line is 180, then GAE = = 60 and GEA = = 100. G x o 60 o 10 o 80 o B A E L 100 o Since the sum of the angles in a triangle is 180, AGE = 180 GAE GEA = = 0. Since AGE = 0, then the reflex angle at G is = 340, so x = 340. Answer: (A) Since the numerators of the five fractions are the same, the largest fraction will be the one with the smallest denominator. To get the smallest denominator, we must subtract (not add) the largest quantity from. 4 Therefore, the largest fraction is 1.

169 005 Gauss Contest s Page 13 We evaluate each of the choices: The largest of the five fractions is = = = = = 4 3 = = = 1 5 =.4 = = Answer: (E) 15. We first try x = 1 in each of the possibilities and we see that it checks for all five possibilities. Next, we try x =. When x =, y = x = =.5. When x =, y = 1.5x = 1.5() = 3. When x =, y = 0.5x + 1 = 0.5() + 1 =. When x =, y = x 0.5 = () 0.5 = 3.5. When x =, y = x = = 4.5. So only the second choice agrees with the table when x =. When x = 3, the second choice gives y = 1.5x = 1.5(3) = 4.5 and when x = 4, y = 1.5x = 1.5(4) = 6. Therefore, the second choice is the only one which agrees with the table. Answer: (B) 16. If the student were to buy 40 individual tickets, this would cost 40 $1.50 = $ If the student were to buy the tickets in packages of 5, she would need to buy 40 5 = 8 packages, and so this would cost 8 $5.75 = $ Therefore, she would save $60.00 $46.00 = $ Answer: (C) In this solution, we try specific values. This does not guarantee correctness, but it will tell us which answers are wrong. We try setting a = (which is even) and b = 1 (which is odd). Then ab = 1 =, a + b = + (1) = 4, a b = () (1) =, a + b + 1 = = 4, and a b = 1 = 1. Therefore, a b is the only choice which gives an odd answer. Since a is even, then ab is even, since an even integer times any integer is even. Since a is even and b is even (since times any integer is even), then their sum a + b is even. Since times any integer is even, then both a and b are both even, so their difference a b

170 005 Gauss Contest s Page 14 is even (since even minus even is even). Since a is even and b is odd, then a + b is odd, so a + b + 1 is even. Since a is even and b is odd, then a b is odd. Therefore, a b is the only choice which gives an odd answer. Answer: (E) For 100 to divide evenly into N and since 100 = 5, then N must have at least factors of and at least factors of 5. From its prime factorization, N already has enough factors of, so the box must provide factors of 5. The only one of the five choices which has factors of 5 is 75, since 75 = Checking, = = Therefore, the only possible correct answer is 75. Multiplying out the part of N we know, we get N = = = 016. We can then try the five possibilities = which is not divisible by = which is not divisible by = which is divisible by = which is not divisible by = which is not divisible by 100. Therefore, the only possibility which works is 75. Answer: (C) 19. In the diagram, B appears to be about 0.4 and C appears to be about 0.6, so B C should be about = 0.4. Also, A appears to be about 0., so B C is best represented by A. Answer: (A) 0. Label the points A, B, C and D, as shown. Through P, draw a line parallel to DC as shown. The points X and Y are where this line meets AD and BC. From this, we see that AX = Y C = 15 3 = 1. Also, P Y = 14 5 = 9. A B 1 1 P X 5 9 Y D 14 C To calculate the length of the rope, we need to calculate AP and BP, each of which is the hypotenuse of a right-angled triangle.

171 005 Gauss Contest s Page 15 Now, AP = = 169 so AP = 13, and BP = = 5, so BP = 15. Therefore, the required length of rope is or 8 m. Answer: (A) 1. 1 Since the area of the large square is 36, then the side length of the large square is 6. Therefore, the diameter of the circle must be 6 as well, and so its radius is 3. Label the four vertices of the small square as A, B, C, and D. Join A to C and B to D. Since ABCD is a square, then AC and BD are perpendicular, crossing at point O, which by symmetry is the centre of the circle. Therefore, AO = BO = CO = DO = 3, the radius of the circle. A B 6 3 O D C But square ABCD is divided into four identical isosceles right-angled triangles. The area of each of these triangles is 1 bh = 1 (3)(3) = 9 so the area of the square is 4 9 = 18. Rotate the smaller square so that its four corners are at the four points where the circle touches the large square. Next, join the top and bottom points where the large square and circle touch, and join the left and right points. By symmetry, these two lines divide the large square into four sections (each of which is square) of equal area. But the original smaller square occupies exactly one-half of each of these four sections, since each edge of the smaller square is a diagonal of one of these sections. Therefore, the area of the smaller square is exactly one-half of the area of the larger square, or 18. Answer: (E)

172 005 Gauss Contest s Page Since there were 50 students surveyed in total and 8 played neither hockey nor baseball, then 4 students in total played one game or the other. Since 33 students played hockey and 4 students played baseball, and this totals = 57 students, then there must be 15 students who are double-counted, that is who play both sports. Let x be the number of students who play both hockey and baseball. Then the number of students who play just hockey is 33 x and the number of students who play just baseball is 4 x. But the total number of students (which is 50) is the sum of the numbers of students who play neither sport, who play just hockey, who play just baseball, and who play both sports. In other words, 8 + (33 x) + (4 x) + x = x + x = x = = x x = 15 Therefore, the number of students who play both sports is 15. Answer: (D) 3. We begin by considering a point P which is where the circle first touches a line L. L C C P P If a circle makes one complete revolution, the point P moves to P and the distance P P is the circumference of the circle, or π m. If we now complete the rectangle, we can see that the distance the centre travels is CC which is exactly equal to P P or π m. Answer: (C) 4. Let the three positive integers be x, y and z. From the given information (x + y) z = 14 and (x y) + z = 14. Let us look at the third number, z, first. From the first equation (x + y) z = 14, we see that z must be a factor of 14. Therefore, the only possible values of z are 1,, 7 and 14. If z = 1, then x + y = 14 and xy = 13. From this, x = 1 and y = 13 or x = 13 and y = 1. (We can find this by seeing that xy = 13 so the only possible values of x and y are 1 and 13, and then checking the first equation to see if they work.) If z =, then x + y = 7 and xy = 1. From this, x = 3 and y = 4 or x = 4 and y = 3.

173 005 Gauss Contest s Page 17 (We can find these by looking at pairs of positive integers which add to 7 and checking if they multiply to 1.) If z = 7, then x + y = and xy = 7. There are no possibilities for x and y here. (Since x and y are positive integers and x + y = then we must have x = 1 and y = 1, but this doesn t work in the second equation.) If z = 14, then x + y = 1 and xy = 0. There are no possibilities for x and y here. (This is because one of them must be 0, which is not a positive integer.) Therefore, the four possible values for x are 1, 13, 3 and 4. Answer: (B) 5. Suppose that there are n coins in the purse to begin with. Since the average value of the coins is 17 cents, then the total value of the coins is 17n. When one coin is removed, there are n 1 coins. Since the new average value of the coins is 18 cents, then the new total value of the coins is 18(n 1). Since 1 penny was removed, the total value was decreased by 1 cent, or 17n 1 = 18(n 1) or 17n 1 = 18n 18 or n = 17. Therefore, the original collection of coins has 17 coins worth 89 cents. Since the value of each type of coin except the penny is divisible by 5, then there must be at least 4 pennies. Removing these pennies, we have 13 coins worth 85 cents. These coins may include pennies, nickels, dimes and quarters. How many quarters can there be in this collection of 13 coins? 1 quarters are worth 1 5 = 300 cents, so the number of quarters is fewer than 1. Could there be as few as 10 quarters? If there were 10 quarters, then the value of the quarters is 50 cents, so the remaining 3 coins (which are pennies, nickels and dimes) are worth 35 cents. But these three coins can be worth no more than 30 cents, so this is impossible. In a similar way, we can see that there cannot be fewer than 10 quarters. Therefore, there are 11 quarters in the collection, which are worth 75 cents. Thus, the remaining coins are worth 10 cents, so must both be nickels. Therefore, the original collection of coins consisted of 11 quarters, nickels and 4 pennies. Answer: (A)

174 Canadian Mathematics Competition An activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 004 s Gauss Contest (Grades 7 and 8) C.M.C. Sponsors: C.M.C. Supporters: Canadian Institute of Actuaries Chartered Accountants Great West Life and London Life Sybase Inc. (Waterloo) ianywhere s 004 Waterloo Mathematics Foundation